Differential Equations: Class 12 Maths Chapter 9 Hinglish mein

Rates of change ki equations, aur woh functions jo unhe satisfy karte hain

Mathematics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
1 simulation
47 practice questions
23 MCQs

Is chapter mein aap kya seekhenge

Differential equation ko pehchaanna aur uska order nikaalna, aur jab defined ho to uski degree bhi
Substitution se check karna ki koi diya hua function, explicit ho ya implicit, differential equation ko solve karta hai ya nahi
General solution aur particular solution mein farq samajhna, aur di gayi conditions se arbitrary constants fix karna
First order, first degree ki aisi equations solve karna jinke variables separate ho jaate hain
Homogeneous equations ko pehchaanna aur unhe substitution y = vx ya x = vy se solve karna
First order linear equations ko integrating factor se solve karna, x mein ya y mein
Slope condition, growth law ya cooling law ko differential equation mein badalna aur use solve karna

NCERT Mathematics Class 12, Chapter 9, pp. 300–337 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

9.1 Introduction

90 °C par daali gayi chai ka glass ek steady pace se thanda nahi hota. Pehle minute mein woh lagbhag chhe degree khota hai; das minute baad, lagbhag 50 °C par, woh ek minute mein sirf do degree khota hai. Jo cheez hum seedhe observe kar sakte hain, woh hai rate: chai room temperature se jitni zyada garam hogi, utni tezi se thandi hogi (Fig. 9.1).

30 °C wale room mein $t$ minute baad temperature ko $T$ likho. Tab observation ban jaata hai $\frac{d T}{d t} = - k (T - 30),$ jahan $k$ ek positive constant hai. Yahan unknown koi number nahi hai, balki ek poora function hai, $T (t),$ aur equation sirf itna batati hai ki woh function kitni tezi se badalta hai. Aisi equation ko differential equation kehte hain.

3/3Slope room se upar wale gap ke proportional hai: $\frac{d T}{d t} = - k (T - 30) .$

Fig. 9.1Chai jitni garam, graph utna steep: cooling ka rate khud temperature par depend karta hai.

Temperature T (degrees Celsius mein) aur time (minutes mein, 0 se 30 minute tak) ka graph. Ek orange curve 90 degree se shuru hoti hai aur 30 degree par bani dashed horizontal line ki taraf mudti hai, jis par room likha hai. Pehle step mein curve ke shuru mein ek steep tangent banaya gaya hai, jis par likha hai 90 degree: steep, tez thandi. Doosre step mein ek doosra, dashed tangent curve ko 50 degree par, lagbhag 11 minute ke paas, chhoota hai; yeh kaafi flat hai aur us par halka slope likha hai. Teesre step mein dono tangents bane rehte hain aur rule dikhta hai: dT by dt barabar minus k times T minus 30.

Is tarah ki equations har us jagah aati hain jahan koi cheez aise rate se badalti hai jise hum describe kar sakte hain: purani lakdi mein carbon-14 ka decay (jisse archaeologists uski age nikaalte hain), kisi town mein rumour ya infection ka phailna, capacitor ka charge hona, ya aisa loan jiska interest continuously compound hota hai. Sabse simple case tum pehle hi solve kar chuke ho. Integrals wale chapter mein $\frac{d y}{d x} = g (x)$ se $y$ nikaalna bas antidifferentiation tha. Nayi baat yeh hai ki rate khud unknown par depend kar sakta hai, jaise chai ke case mein hota hai.

Yeh chapter pehle vocabulary fix karta hai: order, degree, aur general aur particular solutions. Phir first order equations ke liye teen exact methods develop karta hai: variables ko separate karna, homogeneous equations ke liye ek substitution, aur linear equations ke liye integrating factor. Har stage par hum answer ko usi equation mein wapas daal kar check karte hain jisse woh aaya tha.

Quick check

2 mein se 1

Inmein se kaun si differential equations hain?

\frac{d y}{d x} = 3 x + y

x^{2} + y^{2} = 25

\frac{d^{2} s}{d t^{2}} = - 9.8

y = e^{2 x} + 1

Saare sahi options chuno

Answer dekho

Answer

(A) $\frac{d y}{d x} = 3 x + y$
(C) $\frac{d^{2} s}{d t^{2}} = - 9.8$

A aur C mein in unknown functions ke derivatives hain:

y (x)

aur

s (t) .

Circle

x^{2} + y^{2} = 25

aur formula

y = e^{2 x} + 1

mein koi derivative nahi hai; doosra to asal mein is equation ka ek solution hai,

\frac{d y}{d x} = 2 y - 2,

khud aisi equation nahi.

9.2 Basic Concepts

School algebra mein unknown ek number hota hai: $x^{2} - 5 x + 6 = 0$ sirf $x = 2$ aur $x = 3$ ke liye sahi hai, aur kisi ke liye nahi. $x \frac{d y}{d x} + 2 y = 0$ mein unknown ek function hai, $y (x),$ aur equation mein uska derivative aata hai. Variable $x$ independent variable hai aur $y,$ jo us par depend karta hai, dependent variable hai.

Definition

Differential equation

Aisi equation jismein ek unknown function apne kam se kam ek derivative ke saath aata hai.

Example $\frac{d y}{d x} = x^{2} - y,$ jahan $y$ ek unknown function hai jiska variable hai $x .$

Jab sirf ek independent variable ho, jaise is chapter ki har equation mein hai, to humare paas ordinary differential equation hoti hai. Garam ki gayi iron rod ke saath-saath temperature rod par position aur time, dono par depend karta hai, isliye uski equation mein dono ke respect mein partial derivatives aate hain; aisi partial differential equations aage ke courses mein aati hain. Ab se "differential equation" ka matlab ordinary differential equation hai.

Derivatives ke liye hum short forms use karte hain: $y^{'} = \frac{d y}{d x},$ $y^{''} = \frac{d^{2} y}{d x^{2}}$ aur $y^{'''} = \frac{d^{3} y}{d x^{3}} .$ Teesre derivative ke baad dashes bahut ho jaate hain, isliye $n$ th derivative $\frac{d^{n} y}{d x^{n}}$ ko aise likhte hain: $y_{n}$ ya $y^{(n)} .$

9.2.1 Order of a differential equation

Definition

Order

Har term mein gino ki dependent variable ko kitni baar differentiate kiya gaya hai; poori equation mein sabse bada count hi uska order hai.

Example $y^{''} + 4 y = 0$ ka order 2 hai.

Yahan sirf yeh matter karta hai ki $y$ ko kitni baar differentiate kiya gaya. Powers aur coefficients ka koi role nahi:

$\frac{d y}{d x} = 3 x^{2} - y$ ka order 1 hai;
$\frac{d^{2} y}{d x^{2}} + 4 y = \sin x$ ka order 2 hai;
${(\frac{d y}{d x})}^{3} + \frac{d^{3} y}{d x^{3}} = x$ ka order 3 hai, bhale hi cube first derivative par laga hai.

9.2.2 Degree of a differential equation

Degree ke saath ek condition aati hai. Har derivative ( $y^{'},$ $y^{''},$ …) ko equation mein waise hi aana chahiye jaise koi variable polynomial mein aata hai: whole-number powers, sums aur products ke through, kabhi bhi sine, logarithm, exponential ya root ke andar nahi. Doosri taraf, unke coefficients koi bhi functions ho sakte hain, jinke variables hon $x$ aur $y .$

Definition

Degree

Highest order derivative ko dekho. Agar equation apne saare derivatives mein polynomial hai, to degree us derivative par lagi sabse badi whole-number power hai; agar nahi hai, to equation ki koi degree nahi hoti.

Example $(y^{''})^{2} + y^{'} = x$ ka order 2 aur degree 2 hai.

Equation ko do steps mein padho (Fig. 9.2). Pehle highest derivative dhoondho: usse order milta hai. Phir sirf usi derivative ki power dekho: usse degree milti hai. Lower derivatives ki powers, chahe kitni bhi badi hon, koi matter nahi karti.

4/4Power 5 ek lower derivative ki hai, isliye woh degree decide nahi karti.

Fig. 9.2Order highest derivative se; degree usi derivative ki power se.

Ek row mein teen boxes equation ke terms rakhte hain: y ka second derivative cubed, plus x times first derivative ki fifth power, minus y, barabar 0. Pehle step mein ek note kehta hai: pehle highest derivative dhoondho. Doosre step mein pehla box highlight hota hai aur ek arrow order 2, second derivative shabdon ki taraf point karta hai. Teesre step mein us box par lage cube par circle banta hai aur degree 3, yeh power shabd dikhte hain. Chauthe step mein beech wale box par dashed outline aati hai aur ek arrow jis par likha hai power 5: chhodo, lower derivative.

$(y^{'})^{2} + y^{'} \cos y = x ⁣ :$ order 1, degree 2. $\cos y$ se koi dikkat nahi, kyunki $y$ khud derivative nahi hai.
$y^{''} + e^{y^{'}} = 0 ⁣ :$ order 2, lekin $e^{y^{'}}$ koi polynomial nahi hai jiska variable ho $y^{'},$ isliye degree defined nahi hai.
$\sqrt{x} y^{'''} + x^{2} y = 0 ⁣ :$ order 3, degree 1. $x$ ka root chalta hai; kisi derivative ka root nahi chalta.

Example 1NCERT Example 1 · p. 303

Teen equations ka order aur degree

Order aur degree, agar defined ho, nikaalo: (i) ${(\frac{d y}{d x})}^{2} - 4 x y = 0;$ (ii) $\frac{d^{2} y}{d x^{2}} + x^{2} {(\frac{d y}{d x})}^{3} + y = 0;$ (iii) $y^{'''} + x \sin (y^{''}) = 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 4 questions

Practice 9.1: answers ke saath

1
Short answerEasyNCERT Exercise 9.1, Q1–Q5
Har equation ka order aur degree nikaalo. Number type karo, ya jab degree exist na kare to not defined likho.
1. $y^{'''} - e^{y^{''}} = 0 ⁣ :$ order aur degree
  Solution dikhao
  Highest derivative hai $y^{'''},$ isliye order 3 hai. Kyunki $e^{y^{''}}$ koi polynomial nahi hai jiska variable ho $y^{''},$ degree defined nahi hai.
2. $3 y^{'} = 2 y ⁣ :$ order aur degree
  Solution dikhao
  Sirf $y^{'}$ aata hai, first power mein: order 1, degree 1.
3. ${(\frac{d r}{d t})}^{3} - 2 r \frac{d^{2} r}{d t^{2}} = 0 ⁣ :$ order aur degree
  Solution dikhao
  Yahan $r$ ek function hai jiska variable hai $t .$ Highest derivative, $\frac{d^{2} r}{d t^{2}},$ first power mein aata hai: order 2, degree 1. Cube lower derivative par hai.
4. ${(\frac{d^{3} y}{d x^{3}})}^{2} + \sin (\frac{d^{2} y}{d x^{2}}) = 0 ⁣ :$ order aur degree
  Solution dikhao
  Order 3 hai. $y^{''}$ ka sine kisi derivative mein polynomial nahi hai, isliye degree defined nahi hai, bhale hi highest derivative khud sirf squared hai.
5. $\frac{d^{3} y}{d x^{3}} = e^{2 x} + \sin x ⁣ :$ order aur degree
  Solution dikhao
  Sirf $x$ ke functions degree par asar nahi daalte. Order 3, degree 1.
2
Short answerEasyNCERT Exercise 9.1, Q6–Q10
Har equation ka order aur degree nikaalo, Question 1 ki tarah.
1. $(y^{'''})^{2} + (y^{''})^{4} - (y^{'})^{5} + y^{3} = 0 ⁣ :$ order aur degree
  Solution dikhao
  Highest derivative $y^{'''}$ squared hai: order 3, degree 2. Powers 4 aur 5 lower derivatives ki hain.
2. $y^{(4)} - 3 y^{''} + 5 y = 0 ⁣ :$ order aur degree
  Solution dikhao
  Fourth derivative first power mein aata hai: order 4, degree 1.
3. $y^{'} - 2 y = \cos x ⁣ :$ order aur degree
  Solution dikhao
  Order 1, degree 1; $\cos x$ mein koi derivative nahi hai.
4. $2 y^{''} + (y^{'})^{3} = x y ⁣ :$ order aur degree
  Solution dikhao
  Highest derivative $y^{''}$ first power mein aata hai: order 2, degree 1.
5. $y^{''} + e^{y} y^{'} = 0 ⁣ :$ order aur degree
  Solution dikhao
  Factor $e^{y}$ mein koi derivative nahi hai, isliye equation in derivatives mein polynomial hai: $y^{''}$ aur $y^{'} ⁣ :$ order 2, degree 1.
Quick answerEasyNCERT Exercise 9.1, Q11
${(\frac{d^{3} y}{d x^{3}})}^{2} - {(\frac{d y}{d x})}^{4} + e^{d y / d x} = 0$ ki degree hai
2defined nahi43
Answer dekho
Answer (B) defined nahi
Term $e^{d y / d x}$ kisi derivative mein polynomial nahi hai, isliye koi degree exist nahi karti, bhale hi highest derivative sirf squared hai.
Quick answerEasyNCERT Exercise 9.1, Q12
${(\frac{d y}{d x})}^{4} + x^{2} \frac{d^{2} y}{d x^{2}} = 3 y$ ka order hai
412defined nahi
Answer dekho
Answer (C) 2
Highest derivative hai $\frac{d^{2} y}{d x^{2}},$ isliye order 2 hai. 4 ek power hai, order nahi, aur order hamesha defined hota hai.

9.3 General and Particular Solutions of a Differential Equation

$x^{2} - 9 = 0$ ka solution woh number hai jo dono sides ko barabar kar de. Differential equation ka solution ek function hai jo yahi kaam karta hai: use aur uske derivatives ko equation mein daalo, to equation ek identity ban jaati hai, jo kisi interval ke har $x$ ke liye sahi hoti hai.

Seedhi si equation lo, $\frac{d y}{d x} = x .$ Integrate karne par har function $y = \frac{x^{2}}{2} + C$ kaam karta hai, chahe hum constant $C$ kuch bhi chunein. Inke graphs ek hi parabola hain jo upar-neeche khiska diya gaya hai, aur har ek equation ka ek solution curve (jise integral curve bhi kehte hain) hai (Fig. 9.3).

4/4Sirf $C = - 1$ wala member is point se guzarta hai: $(2, 1) .$

Fig. 9.3Family

y = \frac{x^{2}}{2} + C

ka har member slope marks ke saath chalta hai; ek diya hua point ek member ko chun leta hai.

Chhote line segments ka ek square grid, yaani slope marks: har point par segment ka slope x-coordinate ke barabar hai, y-axis par flat aur left ya right jaane par zyada steep. Pehle step mein sirf marks dikhte hain. Doosre step mein ek blue parabola, y equals x squared over 2, marks ke saath chalta hai. Teesre step mein chaar aur parabolas aate hain, upar aur neeche shift kiye hue, jinke lowest points par C equals minus 2 se 2 tak likha hai. Chauthe step mein C equals minus 1 wala parabola ek dashed orange line se alag dikhaya gaya hai, jo marked point (2, 1) se guzarta hai.

Chhote segments dikhate hain ki equation har point par kaun sa slope maangti hai; poora pattern ek slope field (ya direction field) hai. Solution curve woh curve hai jo har jagah in segments ke saath-saath chalti hai. Ek chune hue point se family ka sirf ek member guzarta hai: $(2, 1)$ ke liye humein chahiye $1 = 2 + C,$ isliye $C = - 1.$

Definition

General solution

Solutions ki poori family, free constants ke saath likhi hui, equation ke order jitne constants. Ise primitive bhi kehte hain.

Example $y = \frac{x^{2}}{2} + C$ , is equation ke liye: $\frac{d y}{d x} = x .$

Definition

Particular solution

Family ka ek member: general solution ke har constant ko ek fixed value di gayi hai, aam taur par di gayi conditions poori karne ke liye.

Example $y = \frac{x^{2}}{2} - 1,$ woh member jo is point se guzarta hai: $(2, 1) .$

Order jitne hi constants kyun? $\frac{d^{2} y}{d x^{2}} = 0$ jaisi second order equation ko do baar integrate karna padta hai: pehle $\frac{d y}{d x} = A,$ phir $y = A x + B .$ Do integrations do arbitrary constants laate hain, aur family mein har non-vertical straight line aa jaati hai. Ek line ko alag chunne ke liye humein do facts chahiye, jaise ek point aur ek slope, ya do points.

InteractiveSlope field explorer

Ek square graph jo chuni gayi differential equation ke chhote slope marks se bhara hai, aur paanch tabs: dy by dx equals x; dy by dx equals y over 2; dy by dx equals 3 x squared over 2y; homogeneous equation (2x minus y) dx plus (2y minus x) dy equals 0; aur linear equation dy by dx plus y equals sin x plus cos x. Ek draggable point P, jise do sliders se bhi set kar sakte hain, arbitrary constant fix karta hai; panel P par slope, C ki value aur particular solution dikhata hai, jo ek moti orange curve ki tarah bani hai. Show family button general solution ke doosre members blue mein jodta hai, aur Trace button ek green dot ko uske tangent ke saath curve par chalata hai. Jab tak learner kuch na kare, kuch nahi hilta.

Fig. 9.4

P

ko drag karo (ya sliders use karo) aur dekho ki usse guzarne wala solution kaise badalta hai. Family on karo, ya Trace dabao taaki ek point apne tangent ke saath curve par chale.

Try karke dekho

Slope field par chalo

Tumhe chahiye

ek calculator
graph paper (agar ho)

Pehle guess karo

(0, 1)

se shuru karo aur

x

mein

0.25

ke steps lo, is equation ke slope marks ke saath:

\frac{d y}{d x} = y .

Jab tum pahunchoge

x = 1,

to kya tumhara

y

true value

e \approx 2.72

se neeche hoga, uske barabar hoga ya upar hoga?

neechebilkul barabarupar

Shuru karo yahan se: $x = 0,$ $y = 1.$
Wahan slope hai $\frac{d y}{d x} = y .$ Right mein $0.25$ chalo aur upar $0.25 \times slope .$
Naye point se yahi dohrate raho jab tak pahunch na jao $x = 1,$ aur har naya $y$ table mein likho.
Step ko aadha karke $0.125$ karo aur apne final $y$ ko compare karo is value se: $e .$

Table 9.1

\frac{d y}{d x} = y

ke liye Euler steps

$x$	walk se mila $y$	true value $e^{x}$
0	1	1
0.25		1.284
0.5		1.649
0.75		2.117
1		2.718

Tumhari entries apne aap save hoti hain.

Walk se 1.25, 1.5625, 1.9531 milte hain aur aakhir mein, jab $x = 1,$ lagbhag 2.44, jo neeche hai is value se: $e \approx 2.72.$ $0.125$ ke steps ke saath walk lagbhag 2.57 par khatam hoti hai, jo zyada kareeb hai is value ke: $e .$ Chhote steps solution curve se aur kas kar chipakte hain; yeh step-by-step walk, jise Euler’s method kehte hain, wahi tareeka hai jisse computers aise solutions draw karte hain jinka koi formula nahi hota.

Example 2NCERT Example 2 · p. 305

Solution check karna

Dikhao ki $y = 3 e^{2 x} - 1$ is equation ka solution hai: $\frac{d y}{d x} = 2 y + 2.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3NCERT Example 3 · p. 305

Do constants wali family

Verify karo ki $y = (A + B x) e^{x},$ jahan $A$ aur $B$ constants hain, is equation ko solve karta hai: $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 0.$ Phir woh solution nikaalo jismein $y = 2$ aur $\frac{d y}{d x} = 1$ hon, jab $x = 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 4 questions

Practice 9.2: answers ke saath

1
Short answerEasyNCERT Exercise 9.2, Q1–Q5
Substitution se check karo ki har function apne saamne wali equation ko satisfy karta hai.
1. $y = 2 e^{- x} + 3;$ $y^{''} + y^{'} = 0$
  Solution dikhao
  $y^{'} = - 2 e^{- x}$ aur $y^{''} = 2 e^{- x},$ isliye $y^{''} + y^{'} = 0.$
2. $y = x^{3} - 4 x + C;$ $y^{'} = 3 x^{2} - 4$
  Solution dikhao
  Differentiate karne par milta hai $y^{'} = 3 x^{2} - 4,$ chahe value kuch bhi ho is constant ki: $C .$
3. $y = \sin 2 x + C;$ $y^{'} - 2 \cos 2 x = 0$
  Solution dikhao
  $y^{'} = 2 \cos 2 x,$ isliye $y^{'} - 2 \cos 2 x = 0.$
4. $y = (1 + x^{2})^{3 / 2};$ $(1 + x^{2}) y^{'} = 3 x y$
  Solution dikhao
  Chain rule se $y^{'} = 3 x (1 + x^{2})^{1 / 2} .$ Ise $1 + x^{2}$ se multiply karne par milta hai $3 x (1 + x^{2})^{3 / 2},$ aur yahi hai $3 x y .$
5. $y = B x^{2};$ $x y^{'} = 2 y$
  Solution dikhao
  $y^{'} = 2 B x,$ isliye $x y^{'} = 2 B x^{2} = 2 y .$
2
Short answerMediumNCERT Exercise 9.2, Q6–Q10
Neeche har relation ek function define karta hai, $y,$ kuch sirf implicitly. Check karo ki woh apne saamne wali equation ko satisfy karta hai.
1. $y = x \cos x,$ $0 < x < π;$ $x y^{'} = y - x \sqrt{x^{2} - y^{2}}$
  Solution dikhao
  $y^{'} = \cos x - x \sin x,$ isliye $x y^{'} = y - x^{2} \sin x .$
  Saath hi $x^{2} - y^{2} = x^{2} \sin^{2} x .$ Jab $0 < x < π,$ to $x \sin x > 0,$ isliye $x \sqrt{x^{2} - y^{2}} = x^{2} \sin x,$ jaisa chahiye tha.
2. $x e^{y} = y + C;$ $y^{'} = \frac{e^{y}}{1 - x e^{y}}$ (jahan $x e^{y} \neq 1$ )
  Solution dikhao
  Differentiate karo: $e^{y} + x e^{y} y^{'} = y^{'} .$ $y^{'}$ wale terms ek saath karo: $y^{'} (1 - x e^{y}) = e^{y} .$
3. $y + \sin y = x;$ $(y \cos y - \sin y + x) y^{'} = y$
  Solution dikhao
  Differentiate karne par, $y^{'} (1 + \cos y) = 1.$
  Curve par $x - \sin y = y,$ isliye bracket ban jaata hai $y \cos y + y = y (1 + \cos y) .$ Left side ban jaati hai $y (1 + \cos y) y^{'} = y .$
4. $x - y = \tan^{- 1} y;$ $(y^{2} + 2) y^{'} = 1 + y^{2}$
  Solution dikhao
  Differentiate karo: $1 - y^{'} = \frac{y^{'}}{1 + y^{2}} .$ Is se multiply karo, $1 + y^{2} ⁣ :$ $1 + y^{2} = y^{'} (2 + y^{2}) .$
5. $y = \frac{b}{a} \sqrt{a^{2} - x^{2}},$ $- a < x < a;$ $a^{2} y y^{'} + b^{2} x = 0$
  Solution dikhao
  $y^{'} = - \frac{b}{a} \cdot \frac{x}{\sqrt{a^{2} - x^{2}}},$ isliye $y y^{'} = - \frac{b^{2}}{a^{2}} x .$ $a^{2}$ se multiply karo aur rearrange karo.
Quick answerEasyNCERT Exercise 9.2, Q11
Third order differential equation ke general solution mein kitne arbitrary constants hote hain?
1302
Answer dekho
Answer (B) 3
Har integration ka ek constant, aur third order equation ko teen integrations chahiye.
Quick answerEasyNCERT Exercise 9.2, Q12
Second order differential equation ke particular solution mein kitne arbitrary constants hote hain?
1230
Answer dekho
Answer (D) 0
Particular solution ke saare constants fix ho chuke hote hain, isliye usmein ek bhi nahi hota.

9.4 Methods of Solving First Order, First Degree Differential Equations

Ab se har equation ka order 1 aur degree 1 hai, isliye use aise likh sakte hain: $\frac{d y}{d x} = F (x, y),$ ya variables ke roles badal kar aise: $\frac{d x}{d y} = G (x, y) .$ Koi ek trick in sabko solve nahi karti. Hum teen aisi shapes padhenge jo exactly solve ho sakti hain, aur kisi bhi equation ke saath pehla kaam uski shape pehchaanna hai.

Compare

Teen solvable shapes

Variables separable

$F (x, y) = g (x) h (y)$
$y$ ko $d y$ ke saath aur $x$ ko $d x$ ke saath le jao
dono sides integrate karo

Homogeneous

$F (λ x, λ y) = F (x, y)$
$y = v x$ rakho
nayi equation separate ho jaati hai

Linear

$y^{'} + P y = Q,$ jahan $P,$ $Q$ sirf $x$ mein hain
$e^{\int P d x}$ se multiply karo
ek exact derivative ko integrate karo

9.4.1 Differential equations with variables separable

Sabse aasaan case tab hai jab $F (x, y)$ do factors mein toot jaaye: ek jismein sirf $x$ ho, aur doosra jismein sirf $y .$ Aisi equation ke variables separable hote hain:

\frac{d y}{d x} = g (x) h (y)

(9.1)

jahan

$g (x)$: sirf $x$ ka function
$h (y)$: sirf $y$ ka function

Jahan $h (y) \neq 0,$ wahan $h (y)$ se divide karo aur multiply karo is se, $d x,$ taaki $y$ wali har cheez $d y$ ke saath aur $x$ wali har cheez $d x$ ke saath ho (Fig. 9.5). Phir har side ko integrate karo:

\begin{aligned} \int \frac{d y}{h (y)} & = \int g (x) d x \\ H (y) & = G (x) + C \end{aligned}

(9.2)

jahan

$H$: $\frac{1}{h (y)}$ ka ek antiderivative
$G$: $g (x)$ ka ek antiderivative
$C$: ek arbitrary constant, jo dono integrals ke liye kaafi hai

3/3Har side ko integrate karo; ek constant $C$ kaafi hai.

Fig. 9.5Variables ko sort karo, phir integrate karo: orange

y

-part aur blue

x

-part opposite sides par pahunch jaate hain.

Colour-coded boxes ki teen rows. Pehli row mein dy by dx equals e to the x, ek blue box mein jis par x-part likha hai, times 1 plus y squared, ek orange box mein jis par y-part likha hai. Doosre step mein doosri row aati hai, jis par sort likha hai: dy over 1 plus y squared orange box mein, barabar e to the x dx blue box mein. Teesre step mein teesri row aati hai, jis par integrate likha hai: inverse tan y orange box mein, barabar e to the x blue box mein, plus C.

Is chapter mein $\log$ ka matlab natural logarithm hai, jiska base hai $e,$ jaise integrals wale chapter mein tha.

Example 4NCERT Example 4 · p. 307

Hyperbolas ki ek family

General solution nikaalo: $\frac{d y}{d x} = \frac{x - 2}{y + 3},$ jahan $y \neq - 3.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 308

Ek inverse tangent aata hai

General solution nikaalo: $\frac{d y}{d x} = e^{x} (1 + y^{2}) .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6NCERT Example 6 · p. 308

Constant fix karna

$\frac{d y}{d x} = y^{2} \cos x$ ka woh particular solution nikaalo jismein $y = \frac{1}{2}$ ho jab $x = 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7NCERT Example 7 · p. 308

Diye gaye point se guzarti curve

$(0, 1)$ se guzarne wali woh curve nikaalo jo satisfy karti hai $(x^{2} + 1) d y = 2 x d x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 309

Slope condition se curve tak

Ek curve ke har point $(x, y)$ par tangent ka slope hai $\frac{3 x^{2}}{2 y},$ aur curve is point se guzarti hai: $(2, 3) .$ Uski equation nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 310

Chai peene layak kab hogi?

90 °C ki chai 30 °C wale room mein chhodi gayi hai, aur uska temperature $\frac{d T}{d t} = - k (T - 30)$ follow karta hai, jahan $t$ minutes mein hai. 4 minute baad woh 70 °C tak thandi ho jaati hai. Woh 50 °C tak kab pahunchegi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Kis equation ke variables separable hain?

\frac{d y}{d x} = x + y

\frac{d y}{d x} = e^{x + y}

\frac{d y}{d x} = \sin (x y)

\frac{d y}{d x} = x^{2} + y^{2}

Answer dekho

Answer (B) $\frac{d y}{d x} = e^{x + y}$

e^{x + y} = e^{x} e^{y}

do functions ka product hai, ek

x

ka aur ek jiska variable hai

y .

x + y

x^{2} + y^{2}

jaise sum ko aise product mein nahi toda ja sakta, aur na hi isko:

\sin (x y) .

Practice · 12 questions

Practice 9.3: answers ke saath

1
Short answerEasyNCERT Exercise 9.3, Q1–Q5
General solution nikaalo.
1. $\frac{d y}{d x} = \frac{1 + \cos 2 x}{1 - \cos 2 x}$
  Solution dikhao
  Answer $y = - \cot x - x + C$
  Fraction hai $\frac{2 \cos^{2} x}{2 \sin^{2} x} = \cot^{2} x = {cosec}^{2} x - 1.$ Integrate karne par milta hai $y = - \cot x - x + C .$
2. $\frac{d y}{d x} = x \sqrt{1 - y^{2}},$ $- 1 < y < 1$
  Solution dikhao
  Answer $\sin^{- 1} y = \frac{x^{2}}{2} + C$
  $\frac{d y}{\sqrt{1 - y^{2}}} = x d x,$ isliye $\sin^{- 1} y = \frac{x^{2}}{2} + C .$
3. $\frac{d y}{d x} = 3 - 2 y$
  Solution dikhao
  Answer $y = \frac{3}{2} + A e^{- 2 x}$
  $\frac{d y}{3 - 2 y} = d x$ se milta hai $- \frac{1}{2} \log ∣ 3 - 2 y ∣ = x + C_{1} .$ To $3 - 2 y = B e^{- 2 x},$ aur $y = \frac{3}{2} + A e^{- 2 x}$ jahan $A = - \frac{B}{2} .$
4. $(1 + y^{2}) x d x + (1 + x^{2}) y d y = 0$
  Solution dikhao
  Answer $(1 + x^{2}) (1 + y^{2}) = A$
  Is se divide karo, $(1 + x^{2}) (1 + y^{2}) ⁣ :$ $\frac{x d x}{1 + x^{2}} + \frac{y d y}{1 + y^{2}} = 0.$ Integrate karne par, $\frac{1}{2} \log (1 + x^{2}) + \frac{1}{2} \log (1 + y^{2}) = C,$ jisse milta hai $(1 + x^{2}) (1 + y^{2}) = A .$
5. $(2 + \sin x) d y - \cos x d x = 0$
  Solution dikhao
  Answer $y = \log (2 + \sin x) + C$
  $d y = \frac{\cos x}{2 + \sin x} d x;$ numerator denominator ka derivative hai, aur denominator hamesha positive hai.
2
Short answerMediumNCERT Exercise 9.3, Q6–Q10
General solution nikaalo.
1. $\frac{d y}{d x} = (4 + y^{2}) (1 + 2 x)$
  Solution dikhao
  Answer $\frac{1}{2} \tan^{- 1} \frac{y}{2} = x + x^{2} + C$
  $\frac{d y}{4 + y^{2}} = (1 + 2 x) d x,$ aur $\int \frac{d y}{4 + y^{2}} = \frac{1}{2} \tan^{- 1} \frac{y}{2} .$
2. $x \log x d y - y d x = 0,$ $x > 1$
  Solution dikhao
  Answer $y = C \log x$
  $\frac{d y}{y} = \frac{d x}{x \log x} .$ $u = \log x$ lene par right side ban jaati hai $\frac{d u}{u},$ isliye $\log ∣ y ∣ = \log (\log x) + \log ∣ C ∣ .$
3. $x^{3} \frac{d y}{d x} = - y^{3}$
  Solution dikhao
  Answer $\frac{1}{x^{2}} + \frac{1}{y^{2}} = C$
  $\frac{d y}{y^{3}} = - \frac{d x}{x^{3}} .$ Integrate karne par, $- \frac{1}{2 y^{2}} = \frac{1}{2 x^{2}} + C_{1};$ ab multiply karo is se: $- 2.$
4. $\frac{d y}{d x} = \tan^{- 1} x$
  Solution dikhao
  Answer $y = x \tan^{- 1} x - \frac{1}{2} \log (1 + x^{2}) + C$
  By parts, jahan $u = \tan^{- 1} x ⁣ :$ $\int \tan^{- 1} x d x = x \tan^{- 1} x - \int \frac{x d x}{1 + x^{2}} .$
5. $3 e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0$
  Solution dikhao
  Answer $\tan y = C (e^{x} - 2)^{3}$
  Is se divide karo, $(2 - e^{x}) \tan y ⁣ :$ $\frac{\sec^{2} y}{\tan y} d y = \frac{3 e^{x}}{e^{x} - 2} d x .$ Dono sides ki form hai $\frac{f^{'}}{f},$ isliye $\log ∣ \tan y ∣ = 3 \log ∣ e^{x} - 2 ∣ + \log ∣ C ∣ .$
3
Short answerMediumNCERT Exercise 9.3, Q11–Q14
Di gayi condition se har equation ka ek solution chuno.
1. $(x^{3} + 2 x^{2} + x + 2) \frac{d y}{d x} = 3 x^{2} + 5 x + 3;$ $y = 0$ jab $x = 0$
  Solution dikhao
  Answer $y = \tan^{- 1} x$ $+ \log \frac{(x + 2) (x^{2} + 1)}{2}$
  Bracket ke factors hain $(x + 2) (x^{2} + 1),$ aur partial fractions se milta hai $\frac{1}{x + 2} + \frac{2 x + 1}{x^{2} + 1} .$
  Integrate karne par, $y = \log ∣ x + 2 ∣ + \log (x^{2} + 1) + \tan^{- 1} x + C .$ Jab $x = 0,$ to $0 = \log 2 + C .$
2. $x (x^{2} - 4) \frac{d y}{d x} = 8;$ $y = 0$ jab $x = 4$
  Solution dikhao
  Answer $y = \log \frac{4 (x^{2} - 4)}{3 x^{2}}$
  Partial fractions se: $\frac{8}{x (x - 2) (x + 2)} = - \frac{2}{x} + \frac{1}{x - 2} + \frac{1}{x + 2} .$
  To $y = \log ∣ \frac{x^{2} - 4}{x^{2}} ∣ + C .$ Jab $x = 4 ⁣ :$ $0 = \log \frac{3}{4} + C,$ isliye $C = \log \frac{4}{3} .$
3. $e^{d y / d x} = x + 1;$ $y = 2$ jab $x = 0$
  Solution dikhao
  Answer $y = (x + 1) \log (x + 1) - x + 2$
  Log lene par, $\frac{d y}{d x} = \log (x + 1) .$ By parts, $\int \log (x + 1) d x = (x + 1) \log (x + 1) - x + C,$ aur $x = 0$ se milta hai $C = 2.$
4. $\frac{d y}{d x} = y \cot x;$ $y = 2$ jab $x = \frac{π}{2}$
  Solution dikhao
  Answer $y = 2 \sin x$
  $\frac{d y}{y} = \cot x d x$ se milta hai $\log ∣ y ∣ = \log ∣ \sin x ∣ + \log ∣ C ∣,$ isliye $y = C \sin x,$ aur condition se milta hai $C = 2.$
Short answerMediumNCERT Exercise 9.3, Q15
$(0, 1)$ se guzarne wali kaun si curve ke har point par $y^{'} = e^{x} \cos x$ hai?
Solution dikhao
Answer $y = \frac{e^{x} (\sin x + \cos x) + 1}{2}$
$\int e^{x} \cos x d x = \frac{e^{x} (\sin x + \cos x)}{2} + C .$
Is point par, $(0, 1) ⁣ :$ $1 = \frac{1}{2} + C,$ isliye $C = \frac{1}{2} .$
Short answerMediumNCERT Exercise 9.3, Q16
Equation $(x + 1) \frac{d y}{d x} = x (y - 1),$ ke liye woh solution curve nikaalo jo is point se guzarti hai: $(0, 3) .$
Solution dikhao
Answer $y = 1 + \frac{2 e^{x}}{x + 1}$
$\frac{d y}{y - 1} = (1 - \frac{1}{x + 1}) d x,$ isliye $\log ∣ y - 1 ∣ = x - \log ∣ x + 1 ∣ + C_{1} .$
Isliye $(y - 1) (x + 1) = A e^{x},$ aur $(0, 3)$ se milta hai $A = 2.$
Short answerEasyNCERT Exercise 9.3, Q17
Ek curve is point se guzarti hai, $(0, 3),$ aur uske har point par tangent ka slope coordinates ke product ke barabar hai. Curve nikaalo.
Solution dikhao
Answer $y = 3 e^{x^{2} / 2}$
$\frac{d y}{d x} = x y,$ isliye $\frac{d y}{y} = x d x$ aur $y = A e^{x^{2} / 2} .$ Is point par, $(0, 3),$ $A = 3.$
Short answerMediumNCERT Exercise 9.3, Q18
$(2, 3)$ se guzarne wali ek curve ki yeh property hai: $P (x, y)$ us par kahin bhi ho, $P$ par tangent us segment se teen guna steep hai jo $A (1, 2)$ ko jodta hai is point se: $P .$ Curve nikaalo.
Solution dikhao
Answer $y = 2 + (x - 1)^{3}$
$\frac{d y}{d x} = \frac{3 (y - 2)}{x - 1},$ isliye $\frac{d y}{y - 2} = \frac{3 d x}{x - 1}$ aur $y - 2 = A (x - 1)^{3} .$ Is point par, $(2, 3),$ $A = 1.$
NumericalMediumNCERT Exercise 9.3, Q19
Ek spherical soap bubble aise phulaya jaata hai ki uska volume constant rate se badhta hai. Shuru mein uska radius 2 cm hai aur 7 second baad 4 cm. Uske radius ko time ke function ki tarah likho, jahan time hai $t,$ seconds mein.
Solution dikhao
Answer $r = 2 (t + 1)^{1 / 3}$ cm
Integrate karne par, $\frac{4}{3} π r^{3} = k t + C .$ Jab $t = 0,$ to $C = \frac{4}{3} π \cdot 8.$
Jab $t = 7 ⁣ :$ $\frac{4}{3} π \cdot 64 = 7 k + \frac{4}{3} π \cdot 8,$ isliye $k = \frac{4}{3} π \cdot 8.$
Is se divide karne par, $\frac{4}{3} π ⁣ :$ $r^{3} = 8 t + 8 = 8 (t + 1) .$
NumericalMediumNCERT Exercise 9.3, Q20
Ek account mein paisa saal mein $r %$ ki rate se continuously badhta hai. ₹5000, 15 saal mein ₹15,000 ban jaate hain. Nikaalo $r .$ ( $\log 3 = 1.0986$ use karo.)
%
Solution dikhao
Answerlagbhag $7.32 %$ saalana
$15000 = 5000 e^{15 r / 100},$ isliye $3 = e^{0.15 r}$ aur $r = \frac{\log 3}{0.15} = \frac{1.0986}{0.15} \approx 7.32.$
NumericalEasyNCERT Exercise 9.3, Q21
₹20,000 saal mein 6% par jama kiye gaye hain, continuously compounded. 5 saal baad unki value kitni hogi? ( $e^{0.3} = 1.3499$ use karo.)
₹
Solution dikhao
Answerlagbhag ₹26,998
Jab $t = 5,$ to $P = 20000 e^{0.3} = 20000 \times 1.3499 \approx 26998.$
NumericalMediumNCERT Exercise 9.3, Q22
Ek yeast culture 50,000 cells se shuru hota hai aur apne current size ke proportional rate se badhta hai. 3 ghante baad woh 20% badh chuka hai. Woh 1,00,000 cells tak kab pahunchega? ( $\log 2 = 0.6931$ aur $\log 1.2 = 0.1823$ use karo.)
ghante
Solution dikhao
Answerlagbhag 11.4 ghante baad
$1.2 = e^{3 k}$ se milta hai $k = \frac{\log 1.2}{3} .$ Phir $2 = e^{k t}$ se milta hai $t = \frac{3 \log 2}{\log 1.2} = \frac{2.0793}{0.1823} \approx 11.4.$
Quick answerEasyNCERT Exercise 9.3, Q23
$\frac{d y}{d x} = e^{x - y}$ ka general solution hai
$e^{x} + e^{y} = C$ $e^{- x} + e^{- y} = C$ $e^{y} - e^{x} = C$ $e^{x} + e^{- y} = C$
Answer dekho
Answer (C) $e^{y} - e^{x} = C$
$e^{x - y} = e^{x} e^{- y},$ isliye $e^{y} d y = e^{x} d x$ aur $e^{y} = e^{x} + C .$

9.4.2 Homogeneous differential equations

Kuch equations jaisi hain waise separate nahi hoti, lekin variable badalne ke baad separable ban jaati hain. Inki sabse common family ek scaling test se pehchaani jaati hai.

$x$ ki jagah $λ x$ aur $y$ ki jagah rakho $λ y,$ jahan ek non-zero constant hai $λ .$ Function $3 x^{2} - x y$ ban jaata hai $λ^{2} (3 x^{2} - x y) .$ $\frac{x + 2 y}{x - y}$ mein $λ$ poori tarah cancel ho jaata hai, aur $e^{y / x}$ bhi nahi badalta. Lekin $x + y^{2}$ ban jaata hai $λ x + λ^{2} y^{2},$ jo $λ$ ki kisi ek power times yeh expression nahi hai: $x + y^{2} .$

Homogeneous function of degree n

\begin{aligned} F (λ x, λ y) \\ = λ^{n} F (x, y) \end{aligned}

(9.3)

Definition

Homogeneous function

Dono variables ko kisi bhi non-zero constant $λ$ se scale karne par function multiply ho jaata hai is se: $λ^{n} ⁣ :$ $F (λ x, λ y) = λ^{n} F (x, y) .$ Exponent $n$ uski degree hai.

Example $3 x^{2} - x y$ ki degree 2 hai; $\frac{x + 2 y}{x - y}$ aur $e^{y / x}$ ki degree 0 hai; $x + y^{2}$ homogeneous nahi hai.

$λ = \frac{1}{x}$ chunne par ( $x \neq 0$ ke liye) dikhta hai ki aisa function kaisa dikhta hai: $F (x, y) = x^{n} F (1, \frac{y}{x}),$ yaani $x$ ki ek power times ek aisa function jo sirf $\frac{y}{x}$ par depend karta hai. Jaise, $3 x^{2} - x y = x^{2} (3 - \frac{y}{x}) .$ Isi tarah, $F (x, y) = y^{n} F (\frac{x}{y}, 1)$ jab $y \neq 0.$

Definition

Homogeneous differential equation

Ise aise likhne par, $\frac{d y}{d x} = F (x, y),$ iski right side ki degree zero hoti hai: $x$ aur $y$ ko scale karne par $F$ nahi badalta, isliye $F$ sirf is par depend karta hai: $\frac{y}{x} .$ Form $\frac{d x}{d y} = h (\frac{x}{y})$ bhi count hoti hai.

Example $\frac{d y}{d x} = \frac{x + 2 y}{x - y},$ jiski right side barabar hai $\frac{1 + 2 (y / x)}{1 - (y / x)} .$

Degree zero ki ek picture hai. Origin se guzarti ek line ke saare points ka ratio same hota hai, $\frac{y}{x},$ isliye homogeneous equation un sab par same slope maangti hai (Fig. 9.6). Iska nateeja yeh hai ki kisi solution curve ko origin ke around bada ya chhota karne par ek aur solution curve milti hai.

4/4Ise origin ke around bada ya chhota karne par baaki solutions milte hain.

Fig. 9.6

\frac{d y}{d x} = \frac{2 x - y}{x - 2 y}

ke liye slope sirf ek cheez par depend karta hai,

\frac{y}{x},

isliye solutions ek doosre ki scaled copies hain.

Axes, jinpar origin se guzarti dashed lines hain. Pehle step mein ek line, y equals x over 3, purple mein bani hai aur us par chhe dots hain, origin ke dono taraf teen-teen, aur har dot par slope mark wahi steep segment hai. Doosre step mein har 15 degree par halki lines aati hain jin par slope marks hain; har line par marks parallel hain lekin line badalne par badal jaate hain. Teesre step mein ek blue ellipse, x squared minus xy plus y squared equals 3, marks ke saath chalta hai. Chauthe step mein usi shape ke teen aur dashed ellipses, chhote aur bade, use gher lete hain.

SimulationRay ke saath scaling

Ek coordinate plane jo dono axes par minus 8 se 8 tak hai, ek fixed origin O ke saath, aur do draggable points, P (3, 1) par aur Q (6, 2) par, jo O se guzarti ek hi line par hain. Distances OP aur OQ bane hain aur naape gaye hain, aur har point ki origin mein mirror image dikhayi gayi hai, taaki learner lambda equals 2 aur minus 1 se scale karke mile points ko compare kar sake aur har ek par homogeneous equation ka slope formula check kar sake.

Fig. 9.7Shuru mein

Q

ki position double hai is point ki:

P,

aur mirror point

P^{'}

asal mein

P

hai, scale kiya hua is se:

- 1.

Teeno ek hi line par hain jo guzarti hai is se:

O .

Try karo

$P$ ko $(3, 1)$ par rehne do aur $Q$ ko yahan: $(6, 2) .$ Dono par $\frac{2 x - y}{x - 2 y}$ nikaalo: dono baar 5 aana chahiye. Kya $O Q$ , $O P$ ka double hai?
Mirror point $P^{'}$ hai $(- 3, - 1),$ yaani scaling jahan $λ = - 1.$ Check karo ki wahan bhi slope 5 hai.
$Q$ ko drag karke le jao $(4, 3),$ yaani line se bahar. Ab $\frac{y}{x}$ badal jaata hai, aur slope ho jaata hai $- 2.5.$

DerivationThe substitution

y = v x

Maan lo $\frac{d y}{d x} = g (\frac{y}{x}) .$ Ratio ko apna naam do, $v = \frac{y}{x},$ taaki $y = v x$ ho, jahan $v$ ek function hai is variable ka: $x .$ Product rule se,

\frac{d y}{d x} = v + x \frac{d v}{d x}

(9.4)

Equation mein substitute karne par milta hai $v + x \frac{d v}{d x} = g (v),$ yaani $x \frac{d v}{d x} = g (v) - v .$ Ab variables separate ho jaate hain:

\begin{aligned} \int \frac{d v}{g (v) - v} \\ = \int \frac{d x}{x} + C \end{aligned}

(9.5)

Integrate karo, phir original variables par wapas aane ke liye $v$ ki jagah $\frac{y}{x}$ rakho.

Example 10NCERT Example 10 · p. 314

Ellipses ki ek family

Check karo ki $(2 x - y) d x + (2 y - x) d y = 0$ homogeneous hai, phir use solve karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 316

Equation ke andar ek tangent

Check karo ki $x \frac{d y}{d x} = y + x \tan \frac{y}{x}$ homogeneous hai, aur iska general solution nikaalo ( $x \neq 0$ ).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12NCERT Example 12 · p. 318

$x = v y$ ka use

Is equation ke liye, $x y d x = (x^{2} + y^{2}) d y,$ confirm karo ki equation homogeneous hai aur woh solution nikaalo jismein $x = 2$ ho jab $y = 1.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13NCERT Example 13 · p. 319

Slope rule se curves

Curves ki ek family ki yeh property hai: kisi bhi point $(x, y)$ par tangent ka slope hai $\frac{2 x y}{x^{2} - y^{2}} .$ Family nikaalo aur batao ki curves kya hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 5 questions

Practice 9.4: answers ke saath

1
Short answerMediumNCERT Exercise 9.4, Q1–Q5
Check karo ki har equation homogeneous hai, phir uska general solution nikaalo.
1. $(x y - x^{2}) d y = y^{2} d x$
  Solution dikhao
  Answer $\frac{y}{x} = \log ∣ y ∣ + C$
  $\frac{d y}{d x} = \frac{y^{2}}{x y - x^{2}} = \frac{v^{2}}{v - 1},$ isliye $x \frac{d v}{d x} = \frac{v^{2}}{v - 1} - v = \frac{v}{v - 1} .$
  Phir $(1 - \frac{1}{v}) d v = \frac{d x}{x}$ aur $v - \log ∣ v ∣ = \log ∣ x ∣ + C .$ Kyunki $\log ∣ v ∣ + \log ∣ x ∣ = \log ∣ y ∣,$ yeh ban jaata hai $\frac{y}{x} = \log ∣ y ∣ + C .$
2. $\frac{d y}{d x} = \frac{y + \sqrt{x y}}{x},$ jahan $x, y > 0$
  Solution dikhao
  Answer $2 \sqrt{\frac{y}{x}} = \log x + C$
  Right side hai $v + \sqrt{v},$ isliye $x \frac{d v}{d x} = \sqrt{v},$ $\frac{d v}{\sqrt{v}} = \frac{d x}{x}$ aur $2 \sqrt{v} = \log x + C .$
3. $(x + 2 y) d x - (2 x - y) d y = 0$
  Solution dikhao
  Answer $2 \tan^{- 1} \frac{y}{x} = \log \sqrt{x^{2} + y^{2}} + C$
  $\frac{d y}{d x} = \frac{1 + 2 v}{2 - v},$ isliye $x \frac{d v}{d x} = \frac{1 + v^{2}}{2 - v}$ aur $\frac{(2 - v) d v}{1 + v^{2}} = \frac{d x}{x} .$
  Integrate karne par, $2 \tan^{- 1} v - \frac{1}{2} \log (1 + v^{2}) = \log ∣ x ∣ + C .$ Kyunki $\frac{1}{2} \log (1 + v^{2}) + \log ∣ x ∣ = \log \sqrt{x^{2} + y^{2}},$ answer mil jaata hai.
4. $(x^{2} + 3 y^{2}) d x - 2 x y d y = 0$
  Solution dikhao
  Answer $x^{2} + y^{2} = C x^{3}$
  $\frac{d y}{d x} = \frac{1 + 3 v^{2}}{2 v},$ isliye $x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} .$ Phir $\log (1 + v^{2}) = \log ∣ x ∣ + \log C,$ yaani $x^{2} + y^{2} = C x^{3} .$
5. $x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2}$
  Solution dikhao
  Answer $\tan^{- 1} \frac{y}{x} = \log ∣ x ∣ + C$
  $v + x \frac{d v}{d x} = 1 + v + v^{2},$ isliye $\frac{d v}{1 + v^{2}} = \frac{d x}{x} .$
2
Short answerMediumNCERT Exercise 9.4, Q6–Q10
Confirm karo ki har equation homogeneous hai aur use solve karo.
1. $x d y - y d x = \sqrt{x^{2} - y^{2}} d x,$ jahan $x > 0,$ $∣ y ∣ < x$
  Solution dikhao
  Answer $\sin^{- 1} \frac{y}{x} = \log x + C$
  $\frac{d y}{d x} = v + \sqrt{1 - v^{2}},$ isliye $\frac{d v}{\sqrt{1 - v^{2}}} = \frac{d x}{x} .$
2. $x \tan \frac{y}{x} d y = (x + y \tan \frac{y}{x}) d x$
  Solution dikhao
  Answer $x \cos \frac{y}{x} = C$
  $\frac{d y}{d x} = \frac{y}{x} + \cot \frac{y}{x} .$ Rakho $y = v x ⁣ :$ $x \frac{d v}{d x} = \cot v,$ isliye $\tan v d v = \frac{d x}{x}$ aur $- \log ∣ \cos v ∣ = \log ∣ x ∣ + C_{1} .$
3. $x \frac{d y}{d x} - y = x e^{y / x}$
  Solution dikhao
  Answer $e^{- y / x} + \log ∣ x ∣ = C$
  $x \frac{d v}{d x} = e^{v},$ isliye $e^{- v} d v = \frac{d x}{x}$ aur $- e^{- v} = \log ∣ x ∣ - C .$
4. $x \frac{d y}{d x} = y + y \log \frac{y}{x},$ jahan $x, y > 0$
  Solution dikhao
  Answer $y = x e^{C x}$
  $x \frac{d v}{d x} = v \log v .$ Rakho $u = \log v,$ to $\frac{d v}{v \log v} = \frac{d u}{u},$ isliye $\log ∣ \log v ∣ = \log ∣ x ∣ + \log ∣ C ∣$ aur $\log \frac{y}{x} = C x .$
5. $y \frac{d x}{d y} = x + y \cos^{2} \frac{x}{y}$
  Solution dikhao
  Answer $\tan \frac{x}{y} = \log ∣ y ∣ + C$
  Rakho $x = v y ⁣ :$ $v + y \frac{d v}{d y} = v + \cos^{2} v,$ isliye $\sec^{2} v d v = \frac{d y}{y} .$
3
Short answerHardNCERT Exercise 9.4, Q11–Q15
Har equation solve karo aur woh solution chuno jo condition poori kare.
1. $(x^{2} - 2 y^{2}) d x + x y d y = 0;$ $y = 2$ jab $x = 1$
  Solution dikhao
  Answer $y^{2} = x^{2} + 3 x^{4}$
  $\frac{d y}{d x} = 2 v - \frac{1}{v},$ isliye $x \frac{d v}{d x} = \frac{v^{2} - 1}{v}$ aur $\frac{v d v}{v^{2} - 1} = \frac{d x}{x} .$
  To $v^{2} - 1 = A x^{2},$ yaani $y^{2} - x^{2} = A x^{4} .$ Is point par, $(1, 2),$ $A = 3.$
2. $x^{2} \frac{d y}{d x} = y^{2} + 2 x y;$ $y = 1$ jab $x = 1$
  Solution dikhao
  Answer $y = \frac{x^{2}}{2 - x}$
  $x \frac{d v}{d x} = v^{2} + v .$ Partial fractions, $\frac{1}{v (v + 1)} = \frac{1}{v} - \frac{1}{v + 1},$ se milta hai $\frac{v}{v + 1} = A x,$ yaani $\frac{y}{x + y} = A x .$
  Is point par, $(1, 1),$ $A = \frac{1}{2} ⁣ :$ $2 y = x (x + y),$ isliye $y = \frac{x^{2}}{2 - x} .$
3. $x \frac{d y}{d x} = y + x \cos^{2} \frac{y}{x};$ $y = \frac{π}{4}$ jab $x = 1$
  Solution dikhao
  Answer $\tan \frac{y}{x} = 1 + \log ∣ x ∣$
  $x \frac{d v}{d x} = \cos^{2} v,$ isliye $\sec^{2} v d v = \frac{d x}{x}$ aur $\tan v = \log ∣ x ∣ + C .$ Jab $x = 1,$ to $\tan \frac{π}{4} = 1 = C .$
4. $x \frac{d y}{d x} = y + x \cos \frac{y}{x};$ $y = 0$ jab $x = 1$
  Solution dikhao
  Answer $\sec \frac{y}{x} + \tan \frac{y}{x} = x$
  $x \frac{d v}{d x} = \cos v,$ isliye $\sec v d v = \frac{d x}{x}$ aur $\log ∣ \sec v + \tan v ∣ = \log ∣ x ∣ + \log A .$ Is point par, $(1, 0),$ $1 = A .$
5. $x^{3} \frac{d y}{d x} = x^{2} y + y^{3};$ $y = 1$ jab $x = 1$
  Solution dikhao
  Answer $y^{2} = \frac{x^{2}}{1 - 2 \log ∣ x ∣}$
  $x \frac{d v}{d x} = v^{3},$ isliye $\frac{d v}{v^{3}} = \frac{d x}{x}$ aur $- \frac{1}{2 v^{2}} = \log ∣ x ∣ + C .$ Is point par, $(1, 1),$ $C = - \frac{1}{2},$ jisse milta hai $\frac{x^{2}}{y^{2}} = 1 - 2 \log ∣ x ∣ .$
Quick answerEasyNCERT Exercise 9.4, Q16
Substitution ke saath, $y = v x,$ derivative $\frac{d y}{d x}$ ban jaata hai
$x \frac{d v}{d x}$ $v + \frac{d v}{d x}$ $v + x \frac{d v}{d x}$ $v x \frac{d v}{d x}$
Answer dekho
Answer (C) $v + x \frac{d v}{d x}$
Product rule se, $\frac{d}{d x} (v x) = v + x \frac{d v}{d x} .$
Quick answerEasyNCERT Exercise 9.4, Q17
Inmein se kaun si homogeneous differential equation hai?
$(x^{2} - x y) d x + (y^{2} + 3 x y) d y = 0$ $(2 x + 3 y + 1) d y - (x - y) d x = 0$ $(x^{2} + y) d x + x y d y = 0$ $x y d x - (x^{3} - y^{2}) d y = 0$
Answer dekho
Answer (A) $(x^{2} - x y) d x + (y^{2} + 3 x y) d y = 0$
A mein dono brackets degree 2 ke homogeneous hain, isliye $\frac{d y}{d x}$ degree 0 ka ratio hai. Constant 1 B ko bigaad deta hai, aur C aur D alag-alag degrees mix karte hain.

9.4.3 Linear differential equations

First order equation linear hoti hai jab $y$ aur $\frac{d y}{d x}$ sirf first power mein aayein aur kabhi aapas mein multiply na hon. Aisi har equation ko is tarah arrange kar sakte hain

First order linear equation

\frac{d y}{d x} + P y = Q

(9.6)

jahan

$P, Q$: constants ya sirf $x$ ke functions

$\frac{d y}{d x} - 2 y = x,$ jahan $P = - 2$ aur $Q = x;$
$\frac{d y}{d x} + 2 x y = e^{- x^{2}},$ jahan $P = 2 x;$
$x \frac{d y}{d x} + y = \cos x,$ jo ban jaati hai $\frac{d y}{d x} + \frac{1}{x} y = \frac{\cos x}{x}$ jab divide karte hain is se: $x .$

Doosri taraf, $y \frac{d y}{d x} + x = 0$ ( $y$ aur $\frac{d y}{d x}$ ka product) aur $\frac{d y}{d x} + y^{2} = x$ (ek square) linear nahi hain. Variables ke roles badalne par linear form hai $\frac{d x}{d y} + P_{1} x = Q_{1},$ jismein $P_{1}$ aur $Q_{1}$ sirf $y$ par depend karte hain (ya constants hain), jaise $\frac{d x}{d y} + 2 x = y e^{- y} .$

DerivationIntegrating factor dhoondhna

Left side $\frac{d y}{d x} + P y$ lagbhag ek product ka derivative hai. Poori equation ko ek positive function se multiply karo, $g (x),$ jise abhi chunna baaki hai:

g \frac{d y}{d x} + P g y = Q g

Product rule se, $\frac{d}{d x} (y g) = g \frac{d y}{d x} + g^{'} y .$ Left side isse bilkul match karti hai jab $g^{'} = P g,$ yaani jab $\frac{g^{'}}{g} = P .$ Integrate karne par, $\log g = \int P d x,$ isliye

Integrating factor

g (x) = e^{\int P d x}

(9.7)

Is choice ke saath equation ban jaati hai $\frac{d}{d x} (y g) = Q g,$ aur ek integration se kaam khatam:

\begin{aligned} y e^{\int P d x} \\ = \int Q e^{\int P d x} d x \\ + C \end{aligned}

(9.8)

Definition

Integrating factor

Woh function $e^{\int P d x}$ jo $\frac{d y}{d x} + P y = Q$ ki left side ko ek exact derivative mein badal deta hai, $\frac{d}{d x} (y e^{\int P d x}) .$ Ise I.F. likhte hain.

Example $\frac{d y}{d x} + \frac{2}{x} y = 1$ ke liye, jahan $x > 0,$ I.F. $= e^{2 \log x} = x^{2} .$

Method teen steps mein:

Poori equation ko divide karo taaki $\frac{d y}{d x}$ ka coefficient 1 ho, phir padh lo $P$ aur $Q .$
Nikaalo $e^{\int P d x},$ yaani integrating factor; yahan koi constant nahi chahiye.
$Q$ times integrating factor ko integrate karo, jodo $C,$ aur integrating factor se divide karo; isse milta hai $y,$ jaise (9.8) mein.

Equation $\frac{d x}{d y} + P_{1} x = Q_{1}$ ke liye steps wahi hain, bas letters badal jaate hain:

\begin{aligned} x e^{\int P_{1} d y} \\ = \int Q_{1} e^{\int P_{1} d y} d y \\ + C \end{aligned}

(9.9)

Example 14NCERT Example 14 · p. 324

Har solution ek sine wave ke paas pahunchta hai

General solution nikaalo: $\frac{d y}{d x} + y = \sin x + \cos x .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Term $C e^{- x}$ khatam hota jaata hai, isliye har solution is ke paas pahunchta hai: $y = \sin x .$

Fig. 9.8General solution hai

y = \sin x + C e^{- x} ⁣ :

alag-alag shuruaat, lekin long run mein same behaviour.

Axes jinmein x 0 se lagbhag 7.5 tak hai, pi aur 2 pi par ticks ke saath, aur y minus 3 se 3 tak. Pehle step mein ek moti blue sine wave, y equals sin x, bani hai: C equals 0 wala solution. Doosre step mein chhe dashed orange curves y-axis par heights 3, 2, 1, minus 1, minus 2 aur minus 3 se shuru hoti hain, ends par C equals 3 aur C equals minus 3 likha hai, aur sine wave ki taraf mudti hain. Teesre step mein x equals 3.6 ke aage ka region shaded hai, jahan saari curves wave mein mil chuki hain, aur labels hain milti curves aur C e to the minus x tends to 0.

Example 15NCERT Example 15 · p. 325

Power wala integrating factor

General solution nikaalo: $x \frac{d y}{d x} - 2 y = x^{3},$ jahan $x \neq 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 326

y ki jagah x mein linear

Solve karo $y^{2} d x + (3 x y - 1) d y = 0,$ jahan $y > 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 17NCERT Example 17 · p. 326

Ek particular solution

$\frac{d y}{d x} + y \tan x = \sec x$ ka woh solution nikaalo jismein $y = 2$ ho jab $x = 0,$ jahan $- \frac{π}{2} < x < \frac{π}{2} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 18NCERT Example 18 · p. 327

Slope rule se curve, linear method se

Ek curve origin se guzarti hai, aur har point $(x, y)$ par uska slope abscissa ke square plus ordinate ke barabar hai. Curve nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 7 questions

Practice 9.5: answers ke saath

1
Short answerMediumNCERT Exercise 9.5, Q1–Q6
General solution nikaalo.
1. $\frac{d y}{d x} + 3 y = \cos x$
  Solution dikhao
  Answer $y = \frac{3 \cos x + \sin x}{10} + C e^{- 3 x}$
  I.F. $= e^{3 x},$ aur $y e^{3 x} = \int e^{3 x} \cos x d x = \frac{e^{3 x} (3 \cos x + \sin x)}{10} + C$ (do baar by parts).
2. $\frac{d y}{d x} - 2 y = e^{3 x}$
  Solution dikhao
  Answer $y = e^{3 x} + C e^{2 x}$
  I.F. $= e^{- 2 x},$ aur $y e^{- 2 x} = \int e^{x} d x = e^{x} + C .$
3. $\frac{d y}{d x} + \frac{3 y}{x} = \frac{1}{x^{2}},$ $x > 0$
  Solution dikhao
  Answer $y = \frac{1}{2 x} + \frac{C}{x^{3}}$
  I.F. $= e^{3 \log x} = x^{3},$ aur $x^{3} y = \int x d x = \frac{x^{2}}{2} + C .$
4. $\frac{d y}{d x} + y \cot x = \cos x,$ $0 < x < π$
  Solution dikhao
  Answer $y = \frac{\sin x}{2} + C cosec x$
  I.F. $= e^{\log \sin x} = \sin x,$ aur $y \sin x = \int \sin x \cos x d x = \frac{\sin^{2} x}{2} + C .$
5. $x^{2} \frac{d y}{d x} + y = e^{1 / x},$ $x \neq 0$
  Solution dikhao
  Answer $y = e^{1 / x} (C - \frac{1}{x})$
  Is se divide karo, $x^{2} ⁣ :$ $P = \frac{1}{x^{2}},$ I.F. $= e^{- 1 / x} .$ Phir $y e^{- 1 / x} = \int \frac{d x}{x^{2}} = - \frac{1}{x} + C .$
6. $x \frac{d y}{d x} + y = \log x,$ $x > 0$
  Solution dikhao
  Answer $y = \log x - 1 + \frac{C}{x}$
  Left side pehle se hi hai $\frac{d}{d x} (x y) .$ To $x y = \int \log x d x = x \log x - x + C .$
2
Short answerHardNCERT Exercise 9.5, Q7–Q12
General solution nikaalo.
1. $x \log x \frac{d y}{d x} + y = 2 x \log x,$ $x > 1$
  Solution dikhao
  Answer $y = 2 x - \frac{2 x - C}{\log x}$
  Is se divide karo, $x \log x ⁣ :$ $P = \frac{1}{x \log x},$ I.F. $= e^{\log (\log x)} = \log x .$ Phir $y \log x = \int 2 \log x d x = 2 x \log x - 2 x + C .$
2. $(4 + x^{2}) d y + 2 x y d x = e^{x} d x$
  Solution dikhao
  Answer $y = \frac{e^{x} + C}{4 + x^{2}}$
  Left side hai $d [(4 + x^{2}) y],$ isliye $(4 + x^{2}) y = e^{x} + C .$ (Standard form mein I.F. hai $4 + x^{2}$ .)
3. $x \frac{d y}{d x} + y - x y \tan x = x,$ $0 < x < \frac{π}{2}$
  Solution dikhao
  Answer $y = \tan x + \frac{1}{x} + \frac{C}{x \cos x}$
  Is se divide karo, $x ⁣ :$ $P = \frac{1}{x} - \tan x,$ isliye I.F. $= e^{\log x + \log \cos x} = x \cos x .$ Phir $x y \cos x = \int x \cos x d x = x \sin x + \cos x + C .$
4. $(2 x + y) \frac{d y}{d x} = 1$
  Solution dikhao
  Answer $x = - \frac{y}{2} - \frac{1}{4} + C e^{2 y}$
  $\frac{d x}{d y} - 2 x = y,$ jiska I.F. hai $e^{- 2 y} .$ Phir $x e^{- 2 y} = \int y e^{- 2 y} d y = - (\frac{y}{2} + \frac{1}{4}) e^{- 2 y} + C .$
5. $y d x + (2 x - y^{3}) d y = 0,$ $y > 0$
  Solution dikhao
  Answer $x = \frac{y^{3}}{5} + \frac{C}{y^{2}}$
  $\frac{d x}{d y} + \frac{2}{y} x = y^{2},$ I.F. $= y^{2},$ aur $x y^{2} = \int y^{4} d y = \frac{y^{5}}{5} + C .$
6. $(x + 2 y^{3}) \frac{d y}{d x} = y,$ $y > 0$
  Solution dikhao
  Answer $x = y^{3} + C y$
  $\frac{d x}{d y} - \frac{1}{y} x = 2 y^{2},$ I.F. $= \frac{1}{y},$ aur $\frac{x}{y} = \int 2 y d y = y^{2} + C .$
3
Short answerHardNCERT Exercise 9.5, Q13–Q15
Solve karo, phir condition se constant fix karo.
1. $\frac{d y}{d x} - y \tan x = \cos x;$ $y = 0$ jab $x = 0$
  Solution dikhao
  Answer $y = \frac{x + \sin x \cos x}{2 \cos x}$
  I.F. $= e^{- \int \tan x d x} = \cos x,$ isliye $y \cos x = \int \cos^{2} x d x = \frac{x}{2} + \frac{\sin 2 x}{4} + C,$ aur $x = 0$ se milta hai $C = 0.$
2. $(1 + x^{2}) \frac{d y}{d x} + 4 x y = \frac{1}{1 + x^{2}};$ $y = 1$ jab $x = 0$
  Solution dikhao
  Answer $y = \frac{x + 1}{(1 + x^{2})^{2}}$
  Is se divide karo, $1 + x^{2} ⁣ :$ $P = \frac{4 x}{1 + x^{2}},$ I.F. $= (1 + x^{2})^{2} .$ Phir $y (1 + x^{2})^{2} = \int d x = x + C,$ aur $C = 1.$
3. $\frac{d y}{d x} - 2 y \cot x = \cos x;$ $y = 3$ jab $x = \frac{π}{2}$
  Solution dikhao
  Answer $y = 4 \sin^{2} x - \sin x$
  I.F. $= e^{- 2 \log \sin x} = \frac{1}{\sin^{2} x},$ aur $\frac{y}{\sin^{2} x} = \int \frac{\cos x}{\sin^{2} x} d x = - \frac{1}{\sin x} + C .$ To $y = C \sin^{2} x - \sin x,$ aur $x = \frac{π}{2}$ se milta hai $C = 4.$
Short answerMediumNCERT Exercise 9.5, Q16
$(0, 1)$ se guzarne wali woh curve nikaalo jiska har point $(x, y)$ par slope abscissa ke double minus ordinate ke barabar hai.
Solution dikhao
Answer $y = 2 x - 2 + 3 e^{- x}$
$\frac{d y}{d x} + y = 2 x,$ I.F. $e^{x},$ $y e^{x} = \int 2 x e^{x} d x = 2 (x - 1) e^{x} + C .$ To $y = 2 x - 2 + C e^{- x},$ aur $(0, 1)$ se milta hai $C = 3.$
Short answerMediumNCERT Exercise 9.5, Q17
Ek khaas curve ke point $(x, y)$ par tangent ka slope hamesha is se 3 kam hota hai: $y - x .$ Curve is point se guzarti hai, $(0, 5);$ curve nikaalo.
Solution dikhao
Answer $y = x + 4 + e^{x}$
Standard form mein, $\frac{d y}{d x} - y = - x - 3,$ jiska I.F. hai $e^{- x} .$
$y e^{- x} = \int (- x - 3) e^{- x} d x = (x + 4) e^{- x} + C,$ isliye $y = x + 4 + C e^{x} .$ Is point par, $(0, 5),$ $C = 1.$
Quick answerEasyNCERT Exercise 9.5, Q18
$x \frac{d y}{d x} + 2 y = x^{3}$ ( $x > 0$ ) ka integrating factor hai
$x^{2}$ $\frac{1}{x^{2}}$ $2 \log x$ $e^{x^{2}}$
Answer dekho
Answer (A) $x^{2}$
Standard form mein $P = \frac{2}{x},$ aur $e^{\int P d x} = e^{2 \log x} = x^{2} .$
Quick answerMediumNCERT Exercise 9.5, Q19
$(4 - y^{2}) \frac{d x}{d y} - y x = y^{2}$ ( $- 2 < y < 2$ ) ka integrating factor hai
$\frac{1}{\sqrt{4 - y^{2}}}$ $4 - y^{2}$ $\sqrt{4 - y^{2}}$ $e^{- y^{2} / 2}$
Answer dekho
Answer (C) $\sqrt{4 - y^{2}}$
Yahan $P_{1} = - \frac{y}{4 - y^{2}},$ aur $\int P_{1} d y = \frac{1}{2} \log (4 - y^{2}),$ isliye I.F. $= \sqrt{4 - y^{2}} .$

Miscellaneous Examples

Aakhri chaar examples is chapter ki skills ko mix karte hain. Kisi bhi ko shuru karne se pehle tay karo ki tumhare paas kis tarah ki equation hai (Fig. 9.9).

1/4Equation ko aise likho, $\frac{d y}{d x} = F (x, y),$ phir teen sawaal poochho.

Fig. 9.9Kisi bhi first order, first degree equation se poochne ke teen sawaal.

Upar ek box mein dy by dx equals F of x and y likha hai. Uske neeche teen question boxes ek ke neeche ek hain, jo no likhe arrows se jude hain: F equals g of x times h of y?, F of lambda x and lambda y equals F of x and y?, aur y dash plus P y equals Q?. Har ek se yes likha ek arrow right mein ek answer box tak jaata hai: separate, put y equals v x, aur IF lagao. Pehle step mein saari rows ek jaisi dikhti hain. Doosre, teesre aur chauthe step mein baari-baari se separable, homogeneous aur linear rows colour mein highlight hoti hain aur baaki fade ho jaati hain.

Example 19NCERT Example 19 · p. 329

Second order ka check

Verify karo ki $y = A x + \frac{B}{x},$ jahan arbitrary constants hain $A$ aur $B,$ equation $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - y = 0$ ko satisfy karta hai, jab $x \neq 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 330

Logarithm ko kholna

Ek curve origin se guzarti hai, aur us par har point $(x, y)$ par slope ka natural logarithm barabar hai $x - 2 y .$ Curve nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 331

Substitution ke baad ek perfect square

Solve karo $x^{2} \frac{d y}{d x} = x^{2} + 3 x y + y^{2},$ jahan $x \neq 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 22NCERT Example 22 · p. 333

x mein linear, andar ek substitution ke saath

$d x = (\sin y - x) \cos y d y$ ka woh solution nikaalo jismein $x = 0$ ho jab $y = 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Har equation ko us method se match karo jo use solve karta hai.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

$\frac{d y}{d x} = \frac{e^{x}}{y}$ : variables separate karo
$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{x y}$ : rakho $y = v x$
$\frac{d y}{d x} + \frac{y}{x} = e^{x}$ : integrating factor use karo
$\frac{d x}{d y} = \frac{x}{y} + \cos \frac{x}{y}$ : rakho $x = v y$

\frac{e^{x}}{y}

do functions ka product hai, ek

x

ka aur ek jiska variable hai

y;

doosri right side ki degree 0 hai; teesri linear hai is variable mein:

y;

chauthi ek function hai jiska input hai

\frac{x}{y}

aur yeh is form mein aati hai:

\frac{d x}{d y} .

History se

Pehli equations kisne solve ki?

Differential equations calculus ke saath hi paida hui. November 1675 mein Gottfried Wilhelm Leibniz ne apne notes mein $\int y d y = \frac{1}{2} y^{2}$ likha, aur isi ke saath integral sign aur $d y$ notation dono shuru kiye jo hum aaj bhi use karte hain. Bees saal ke andar unhone is chapter ke teeno methods dhoondh liye: 1691 mein separation of variables, uske jaldi baad homogeneous equations ka reduction, aur phir linear equations ka method.

Shabd baad mein aaye. Jakob Bernoulli ne 1690 mein answer ko equation ka integral kaha, aur unke chhote bhai Johann ne separation method ko uska naam diya. Joseph-Louis Lagrange ne 1774 mein solution shabd use kiya, aur lagbhag ek sadi baad Henri Poincaré ne ise standard bana diya. Poincaré ne differential equations ki qualitative study bhi shuru ki: solution curves ki shapes ko describe karna, jaise slope field karta hai, tab bhi jab unka koi formula exist na kare.

Recap

Summary aur key points

Differential equation mein ek unknown function ke derivatives hote hain; ordinary differential equation mein sirf ek independent variable hota hai (Section 9.2).
Order: sabse zyada differentiate kiye gaye term mein unknown ko kitni baar differentiate kiya gaya hai (Section 9.2.1).
Degree: highest derivative par laga exponent, jo tabhi exist karta hai jab equation apne saare derivatives mein polynomial ho. Dono numbers positive integers hote hain (Section 9.2.2).
Solution woh function hai jo equation ko identity bana deta hai; uska graph ek solution curve hai (Section 9.3).
Order- $n$ equation ke general solution mein $n$ arbitrary constants hote hain; di gayi conditions se sabko fix karne par particular solution milta hai (Section 9.3).
Agar $\frac{d y}{d x} = g (x) h (y),$ to separate karo aur integrate karo: $\int \frac{d y}{h (y)} = \int g (x) d x + C$ (Section 9.4.1).
$F (x, y)$ degree $n$ ka homogeneous hai agar $F (λ x, λ y) = λ^{n} F (x, y);$ $\frac{d y}{d x} = F (x, y)$ homogeneous hai jab $F$ ki degree 0 ho (Section 9.4.2).
Homogeneous equation ke liye rakho $y = v x,$ ya $x = v y$ jab equation ho $\frac{d x}{d y} = h (\frac{x}{y});$ nayi equation separate ho jaati hai (Section 9.4.2).
$\frac{d y}{d x} + P y = Q,$ jahan $P$ aur $Q$ sirf is variable ke functions hain, $x,$ linear hai: $y \cdot I.F. = \int Q \cdot I.F. d x + C$ jahan $I.F. = e^{\int P d x}$ (Section 9.4.3).
$\frac{d x}{d y} + P_{1} x = Q_{1}$ ke liye use karo $I.F. = e^{\int P_{1} d y}$ aur $x \cdot I.F. = \int Q_{1} \cdot I.F. d y + C$ (Section 9.4.3).
Solve karne se pehle equation ko classify karo, aur answer ko wapas substitute karke check karo (Miscellaneous Examples).

Flash cards · 8

Key terms

1 / 8

Miscellaneous Exercise on Chapter 9

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 9.1, Q1–Q5	Practice 9.1 · Q1
NCERT Exercise 9.1, Q6–Q10	Practice 9.1 · Q2
NCERT Exercise 9.1, Q11	Practice 9.1 · Q3
NCERT Exercise 9.1, Q12	Practice 9.1 · Q4
NCERT Exercise 9.2, Q1–Q5	Practice 9.2 · Q1
NCERT Exercise 9.2, Q6–Q10	Practice 9.2 · Q2
NCERT Exercise 9.2, Q11	Practice 9.2 · Q3
NCERT Exercise 9.2, Q12	Practice 9.2 · Q4
NCERT Exercise 9.3, Q1–Q5	Practice 9.3 · Q1
NCERT Exercise 9.3, Q6–Q10	Practice 9.3 · Q2
NCERT Exercise 9.3, Q11–Q14	Practice 9.3 · Q3
NCERT Exercise 9.3, Q15	Practice 9.3 · Q4
NCERT Exercise 9.3, Q16	Practice 9.3 · Q5
NCERT Exercise 9.3, Q17	Practice 9.3 · Q6
NCERT Exercise 9.3, Q18	Practice 9.3 · Q7
NCERT Exercise 9.3, Q19	Practice 9.3 · Q8
NCERT Exercise 9.3, Q20	Practice 9.3 · Q9
NCERT Exercise 9.3, Q21	Practice 9.3 · Q10
NCERT Exercise 9.3, Q22	Practice 9.3 · Q11
NCERT Exercise 9.3, Q23	Practice 9.3 · Q12
NCERT Exercise 9.4, Q1–Q5	Practice 9.4 · Q1
NCERT Exercise 9.4, Q6–Q10	Practice 9.4 · Q2
NCERT Exercise 9.4, Q11–Q15	Practice 9.4 · Q3
NCERT Exercise 9.4, Q16	Practice 9.4 · Q4
NCERT Exercise 9.4, Q17	Practice 9.4 · Q5
NCERT Exercise 9.5, Q1–Q6	Practice 9.5 · Q1
NCERT Exercise 9.5, Q7–Q12	Practice 9.5 · Q2
NCERT Exercise 9.5, Q13–Q15	Practice 9.5 · Q3
NCERT Exercise 9.5, Q16	Practice 9.5 · Q4
NCERT Exercise 9.5, Q17	Practice 9.5 · Q5
NCERT Exercise 9.5, Q18	Practice 9.5 · Q6
NCERT Exercise 9.5, Q19	Practice 9.5 · Q7
NCERT Miscellaneous Exercise on Chapter 9, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise on Chapter 9, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise on Chapter 9, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise on Chapter 9, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise on Chapter 9, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise on Chapter 9, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise on Chapter 9, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise on Chapter 9, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise on Chapter 9, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise on Chapter 9, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise on Chapter 9, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise on Chapter 9, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise on Chapter 9, Q13	Miscellaneous exercise · Q13
NCERT Miscellaneous Exercise on Chapter 9, Q14	Miscellaneous exercise · Q14
NCERT Miscellaneous Exercise on Chapter 9, Q15	Miscellaneous exercise · Q15

Quick check: MCQs

9.1 Introduction2 questions
9.2.2 Degree of a differential equation4 questions
9.3 General and Particular Solutions of a Differential Equation4 questions
9.4.1 Differential equations with variables separable3 questions
9.4.2 Homogeneous differential equations3 questions
9.4.3 Linear differential equations4 questions
Miscellaneous Examples3 questions

Glossary · 14

Key terms (English + Hindi)

arbitrary constantस्वेच्छ अचरsvechchh achar: Aisa constant, jaise $C,$ jo general solution mein koi bhi value le sakta hai.Chapter mein pehli baar yahan aaya →

degreeघातghaat: Highest derivative par laga exponent, jab equation apne derivatives mein polynomial ho; warna defined nahi.
differential equationअवकल समीकरणavakal sameekaran: Aisi equation jismein kisi unknown function ke derivatives hote hain.Chapter mein pehli baar yahan aaya →

general solutionव्यापक हलvyaapak hal: Free constants wala solution, har integration ka ek: $n$ constants, jab equation ka order ho $n .$

homogeneous differential equationसमघातीय अवकल समीकरणsamaghaateey avakal sameekaran: $\frac{d y}{d x} = F (x, y)$ jahan $F$ degree zero ka homogeneous ho.
homogeneous functionसमघातीय फलनsamaghaateey phalan: Aisa $F$ jiske liye $F (λ x, λ y) = λ^{n} F (x, y)$ ho, chahe koi bhi non-zero ho $λ;$ $n$ uski degree hai.

integrating factorसमाकलन गुणकsamaakalan gunak: $e^{\int P d x},$ jo linear equation ki left side ko exact derivative bana deta hai.

linear differential equationरैखिक अवकल समीकरणraikhik avakal sameekaran: First order equation jismein $y$ aur $y^{'}$ sirf first power mein aate hain, isliye yeh $y^{'} + P y = Q$ jaisi dikhti hai, jahan $P,$ $Q$ sirf $x$ par depend karte hain.Chapter mein pehli baar yahan aaya →

orderकोटिkoti: Equation mein kahin bhi unknown function ko sabse zyada kitni baar differentiate kiya gaya hai.
ordinary differential equationसाधारण अवकल समीकरणsaadhaaran avakal sameekaran: Aisi differential equation jismein sirf ek independent variable ho.Chapter mein pehli baar yahan aaya →

particular solutionविशिष्ट हलvishisht hal: Arbitrary constants ke bina solution, jo di gayi conditions se fix hota hai.

solutionहलhal: Aisa function jo substitute karne par differential equation ko identity bana deta hai.Chapter mein pehli baar yahan aaya →
solution curveहल वक्रhal vakra: Solution ka graph; yeh har point par wahi slope follow karta hai jo equation maangti hai.Chapter mein pehli baar yahan aaya →

variables separableचर पृथक्करणीयchar prithakkaraneey: Is form ki equation, $\frac{d y}{d x} = g (x) h (y),$ taaki har variable ko uski apni side par le ja sakein.Chapter mein pehli baar yahan aaya →

Differential Equations: Class 12 Maths Chapter 9 Hinglish mein

Is chapter mein aap kya seekhenge

9.1 Introduction

9.2 Basic Concepts

9.2.1 Order of a differential equation

9.2.2 Degree of a differential equation

Practice 9.1: answers ke saath

9.3 General and Particular Solutions of a Differential Equation

Practice 9.2: answers ke saath

9.4 Methods of Solving First Order, First Degree Differential Equations

9.4.1 Differential equations with variables separable

Practice 9.3: answers ke saath

9.4.2 Homogeneous differential equations

Practice 9.4: answers ke saath

9.4.3 Linear differential equations

Practice 9.5: answers ke saath

Miscellaneous Examples

Summary aur key points

Miscellaneous Exercise on Chapter 9

Practice questions answers ke saath (NCERT “Miscellaneous exercise”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye