KC Sinha: Exercise 23.4- Mathematics Solution Class 12 Chapter 23 अवकल समीकरण

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Question 1

निम्नलिखित अवकल समीकरण को हल करे।
[Solve the following differential equations]
1.$\frac{dy}{dx}=(e^x+1)y$
Sol :
$\frac{dy}{dx}=(e^x+1)y$
$\frac{dy}{dx}=(e^x+1)dx$

दोनो तरफ समाकलन लेने पर

$\int \frac{dy}{dx}=\int (e^x+1)dx$

log |y|=e^x+x+c

2.$\frac{dy}{dx}=\frac{x+1}{2-y}$,y≠2
Sol :
$\frac{dy}{dx}=\frac{x+1}{2-y}$

(2-y)dy=(x+1)dx

दोनो तरफ समाकल लेने पर

$\int (2-y)dy=\int (x+1)dx$

$2y-\frac{y^2}{2}=\frac{x^2}{2}+x+k$

$\frac{4y-y^2}{2}=\frac{x^2+2x+2k}{2}$

0=x²+y²+2x-4y+2k

x²+y²+2x-4y+c=0 
जहाँ c=2k

3.$(1+x^2)\tan ^{-1} x\frac{dy}{dx}+y=0$
Sol :
$\frac{dy}{y}=-\frac{dx}{(1+x^2)\tan^{-1}x}$

दोनो तरफ समाकल लेने पर

$\int \frac{dy}{y}=-\int \frac{dx}{(1+x^2)\tan^{-1}x}$

log y=-log |tan^-1 x|+log c

log y+log|tan^-1 x|=log c

log(y tan^-1 x)=log c⇒y tan ^-1 x=c

4.sec²x tany dx+sec²y tan xdy=0
Sol :
sec²y tan xdy=-sec²x tany dx

$\frac{\text{sec}^2 ydy }{\tan y}=\frac{-\text{sec}^2 x}{\tan x}dx$

Integrating both side

$\int \frac{\sec^2 y}{\tan y}dy=-\int \frac{\sec^2 x}{\tan x}dx$

log |tan y|=-log |tan x|+log c

$log |tan x|+\log |\tan y|=\log c$

log|tan x. tan y)|=log c

tan x tan y=c

5.ydx+xdy=xy(dy-dx)

Sol :

ydx+xdy=xydy-xydx

xdy-xydy=-ydx-xydx

x(1-y)dy=-y(1+x)dx

$\frac{(1-y)}{y}dy=-\frac{(1+x)}{x}dx$

$\left(\frac{1}{y}-1\right)dy=-\left(\frac{1}{x}+1\right)dx$

Integrating both side

$\int \left(\frac{1}{y}-1\right)dy=-\int \left(\frac{1}{x}+1\right)dx$

log y-y=-(log x+x)+c

log x+x+log y-y=c

log xy +x-y=c

6.(1-x²)dy+xydy=xy²dx

Sol :

(1+x²)dy=xy²dx-xydx

(1-x²)dy=xy(y-1)dx

$\frac{dy}{y(y-1)}=\frac{x}{1-x^2}dx$

Integrating both side

$\int \frac{dy}{y(y-1)}=\int \frac{x}{1-x^2}dx$

$\int \left[\frac{1}{y-1}-\frac{1}{y}\right]dy=-\frac{1}{2}\int \frac{-2x}{1-x^2}dx$

log |y-1|-log|y|=$-\frac{1}{2}\log|1-x^2|+\log c$

$\log \frac{y-1}{y}+\log (1-x^2)^{\frac{1}{2}}=\log c$

$\log \frac{y-1}{y}\sqrt{1-x^2}=\log c$

$\frac{y-1}{y}\sqrt{1-x^2}=c$

7.tan ydx+sec² y tan xdy=0
Sol :
sec² y tanxdy=-tan ydx

$\frac{\sec ^2 y}{\tan y}dy=\frac{-dx}{\tan x}$

Integrating both side

$\int \frac{\text{sec }^2y}{\tan y}dy=-\int \text{cot x}dx$

log|tan y|=-log|sin x|+log c

log |sin x|+log|tan y|=log c

log|sin x.tan y|=log c

sin x.tan y=c

8.$\text{cosec x}log y \frac{dy}{dx}+x^2y=0$
Sol :
$\text{cosec x}\log y \frac{dy}{dx}=-x^2y$

$\frac{\log y}{y}dy=\frac{-x^2 dx}{\text{cosec x}}$

Integrating both sides

$\int \frac{\log y}{y}dy=-\int x^2\sin x dx$

log y=t

Differentiating w.r.t y

$\frac{1}{y}=\frac{dt}{dy}$ 

$\frac{dy}{y}=dt$

$\int t dt=-2\left[x^2 \int \sin x dx -\int \left\{\frac{d(x^2)}{dx}.\int \sin x dx \right\}dx\right]$

$\frac{1}{2}t^2=-\left[x^2 \cos x-\int 2x \cos x dx \right]$

$\frac{1}{2} (\log y)^2 =+x^2 \cos x-2 \int x \cos x dx$

$\frac{1}{2}(\log y)^2=x^2 \cos x-2 \left[x\int \cos xdx-\int \left\{\frac{dx}{dx}\int \cos x dx\right\}dx\right]$

$\frac{1}{2}(\log y)^2=x^2 \cos x -2\left[x \sin x-\int \sin x dx\right]$

$\frac{1}{2}(\log y)^2=x^2 \cos x-2x \sin x+2(-\cos x)+c$

$\frac{1}{2}(\log y)^2-x^2\cos x+2x\sin x+2\cos x=c$

9.(x²-yx²)dy+(y²+x²y²)dx=0
Sol :
(x²-yx²)dy=-(y²+x²y²)dx
x²(1-y)dy=-y²(1+x²)dx
$\frac{(1-y)dy}{y^2}=-\frac{(1+x^2)}{x^2}dx$
$\left(\frac{1}{y^2}-\frac{1}{y}\right)dy=-\left(\frac{1}{x^2}+1\right)dx$

Integrating both side

$\int \left(\frac{1}{y^2}-\frac{1}{y}\right)dy=-\int \left(\frac{1}{x^2}+1\right)dx$

$-\frac{1}{y}-\log |y|=-\left(-\frac{1}{x}+x\right)+c$

$x-\frac{1}{x}-\frac{1}{y}-\log |y|=c$

10.$\frac{dy}{dx}=(1+x^2)(1+y^2)$
Sol :
$\frac{dy}{1+y^2}=(1+x^2)dx$

Integrating both sides

$\int \frac{dy}{1+y^2}=\int (1+x^2)dx$

$\tan ^{-1} y=x+\frac{x^3}{3}+c$

11. 2sin3ydx+3xcos3ydy=0
Sol :
2sin3ydx+3xcos3ydy=0

3xcos3ydy=-2sin3ydy

Integrating both sides

$\int \frac{\cos 3y}{\sin 3y}dy=-\frac{2}{3}\int \frac{1}{x}dx$

$\frac{\log|\sin 3y|}{3}=\frac{-2}{3}\log x+\log k$

$\frac{\log|\sin 3y|}{3}=\frac{-\log x^2 +3 \log k}{3}$

log|sin 3y|+log x²=3log k

log|x² sin 3y|=log k³

x² sin 3y=k³

x² sin 3y=c ,जहाँ c=k³

12.$\frac{dy}{dx}=\frac{x}{x^2-1}$
Sol :
$dy=\frac{x}{x^2+1}dx$

Integrating both sides

$\int dy=\frac{1}{2}\int \frac{2x}{x^2+1}dx$

$y=\frac{1}{2}\log|x^2+1|+c$

13.$e^x+e^{-x}\frac{dy}{dx}=(e^x-e^{-x})$
Sol :
$dy=\frac{(e^x-e^{-x})}{e^x+e^{-x}}dx$

Integrating both side

$\int dy=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

y=log|e^x+e^-x|+c

14.$(x+2)\frac{dy}{dx}=x^2+4x-9$
Sol :
$dy=\frac{x^2+4x-9}{x+2}dx$

$dy=\frac{(x^2+4x+4)-13}{(x+2)}dx$

$dy=\frac{(x+2)^2-13}{x+2}dx$

$dy=\left[x+2-\frac{13}{x+2}\right]$

Integrating both sides

$\int dy=\int \left(x+2-\frac{13}{x+2}\right)$

$y=\frac{x^2}{2}+2x-13\log |x+2|+c$

15.$\frac{dy}{dx}=\sin ^3 x\cos ^2 x+xe^x$
Sol :
dy=(sin³ x cos² x+xe^x)dx

dy=[sin x(sin² x)cos² x+xe^x]dx$

dy=[sin x(1-cos²x cos² x+xe^x]dx$

dy=[sin x cos² x-sin cos⁴ x+xe^x]dx

Integrating both sides

$\int dy=\int (\sin x \cos^2 x-\sin x \cos^4 x+xe^x)dx$

Putting cosx=t

Differentiating w.r.t x

$-\sin x=\frac{dt}{dx}$

sin xdx=-dt

$y=-\frac{1}{3}\cos ^3 x+\frac{1}{5}\cos ^5 x+x\int e^x dx-\int \left\{\frac{d(x)}{dx}\int e^x dx\right\}dx$

$y=-\frac{1}{3}\cos ^3 x+\frac{1}{5} \cos ^5 x+x e^x -\int e^x dx$

$y=-\frac{1}{3}\cos ^3 x+\frac{1}{5}\cos ^5 x+ xe^x-e^x +c$

16.$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}dx$
Sol :
$dy=\frac{1-\cos x}{1+\cos x}dx$

$dy=\frac{2\sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}dx$

$dy=\tan^{2} \frac{x}{2} dx $

$dy=\left(\sec ^2 \frac{x}{2}-1\right)dx$

Integrating both sides

$\int dy=\int \left(\sec ^2 \frac{x}{2}-1\right)dx$

$y=2\tan \frac{x}{2}-x+c$

17.ylog ydx-xdy=0
Sol :
-xdy=-y log ydx

$\frac{dy}{y log y}=\frac{dx}{x}$

Integrating both sides

$\int \frac{dy}{y log y}=\int \frac{dx}{x}$

log |log y|=log |x|+log c

log |log y|=log |cx|

log y= cx

y=e^cx

18.$\frac{dy}{dx}+y=1,$y≠1
Sol :
$\frac{dy}{dx}+y=1$

$\frac{dy}{dx}=1-y$

dy=(1-y)dx

$\frac{dy}{1-y}=dx$

Integrating both sides

$\int \frac{dy}{1-y}=\int dx$

-log|1-y|=x+c

19.$(x+1)\frac{dy}{dx}=2xy$
Sol :
$\frac{dy}{dx}=\frac{2x}{x+1}dx$

Integrating both sides

$\int \frac{dy}{dx}=2\int \frac{x}{x+1}dx$

$\int \frac{dy}{dx}=2\int \left[1-\frac{1}{x+1}\right]dx$

log |y|=2[x-log|x+1|]+c

20.$x^5\frac{dy}{dx}=-y^5$
Sol :
$\frac{dy}{y^5}=-\frac{dx}{x^5}$

y^-5dy=-x^-5dx

Integrating both sides

$\int y^{-5}dy=-\int x^{-5}dx$

$\frac{y^{-4}}{-4}=-\frac{x^{-4}}{-4}+k$

$-\frac{x^{-4}}{4}-\frac{y^{-4}}{4}=k$

$-\frac{1}{4x^4}-\frac{1}{4y^4}=k$

$\frac{-(y^4+x^4)}{4x^4y^4}=k$

x⁴+y⁴=-4kx⁴y⁴

x⁴+y⁴=cx⁴+y⁴

जहाँ c=-4k

21.$e^x\sqrt{1-y^2}dx+\frac{y}{x}dy=0$
Sol :
$\frac{y}{x}dy=e^x\sqrt{1-y^2}dx$

$\frac{y}{\sqrt{1-y^2}}dy=xe^x dx$

Integrating both sides

$-\frac{1}{2}\int \frac{-2y}{\sqrt{1-y^2}}dy=\int xe^x dx$

$-\frac{1}{2}\times 2 \sqrt{1-y^2}=x\int e^x dx-\int \left\{\frac{dx}{dx}\int e^x dx\right\}dx$

$-\sqrt{1-y^2}+c=xe^x-e^x$

22.x(1+y²)dx-y(1-x²)dy=0
Sol :
-y(1-x²)dy=-x(1+y²)dy=0

$\frac{y}{1+y^2}dy=\frac{x}{1-x^2}dx$

Integrating both sides 

$\frac{1}{2}\int \frac{2y}{1+y^2}dy=\frac{1}{2}\int \frac{-2x}{1-x^2}dx$

$\frac{1}{2}\log |1+y^2|=-\frac{1}{2}\log|1-x^2|+\log c$

$\frac{1}{2}\log(1-x^2)(1+y^2)=\log c$

log(1-x²)(1+y²)=2log c

log(1-x²)(1+y²)=log c²

(1-x²)(1+y²)=c²

(1-x²)(1+y²)=k , जहाँ k=c²

23.$\frac{dy}{dx}=1+x+y+xy$
Sol :
$\frac{dy}{dx}=1(1+x)+y(1+x)$

$\frac{dy}{dx}=(1+x)(1+y)$

$\frac{dy}{1+y}=(1+x)dx$

दोनो तरफ समाकलन लेने पर

$\int \frac{dy}{1+y}=\int (1+x)dx$

log |1+y|=$x+\frac{x^2}{2}+c$

24.e^x tan ydx+(1-e^x)sec² ydy=0
Sol :
(1-e^x)sec² ydy=-e^x tan ydx

$\frac{\sec ^2 y}{\tan y}dy=\frac{-e^x}{1-e^x}dx$

Integrating both sides

$\int \frac{\sec ^2 y}{\tan y}dy=\int \frac{-e^x}{1-e^x}dx$

log|tan y|=log|1-e^x|+log |c|

log|tan y|=log|(1-e^x)c|

tan y=(1-e^x)c

25.$\frac{dy}{dx}=e^{x-y}+x^2 e^{-7}$
Sol :
$\frac{dy}{dx}=e^{-y}(e^x+x^2)$

$\frac{dy}{e^{-y}}=(e^x+x^2)dx$

e^y dy=(e^x+x²)dx

Integrating both sides

$\int e^y dy=\int (e^x+x^2)dx$

$e^y=e^x+\frac{x^3}{3}+c$

26.$\log_e \left(\frac{dy}{dx}\right)=3x-5y $
Sol :
$\frac{dy}{dx}=e^{3x-5y}$

$\frac{dy}{dx}=\frac{e^{3x}}{e^{5y}}$

e^5y dy=e^3x dx

Integrating both sides

$\int e^{5y}dy=\int e^{3x}dx$

$\frac{e^{5y}}{5}=\frac{e^{3x}}{3}+k$

$\frac{e^{5y}}{5}-\frac{e^{3x}}{3}=k$

$\frac{3e^{5y}-5e^{3x}}{15}=k$

3e^5y -5e^3x=15k

3e^5y -5e^3x=15k , जहाँ c=15k

27.$\frac{dy}{dx}=e^{x+y}$

Sol :

$\frac{dy}{dx}=e^x.e^y$

$\frac{dy}{dx}=e^x dx$

e^-ydy=e^x dx

Integrating both sides

$\int e^{-y}dy=\int e^{x}dx$

-e^-ydy=e^x+c

28.x(xdy-ydx)=ydx

Sol :

x²dy-xydx=y dx

x²dy=ydx+xydx

x²dy=y(1+x)dx

$\frac{dy}{y}=\frac{(1+x)}{x^2}dx$

$\frac{dy}{y}=\left(\frac{1}{x^2}+\frac{1}{x}\right)dx$

log |y|$=-\frac{1}{x}+\log|x|+\log|c|$

log|y|$=-\frac{1}{x}+\log|cx|$

log|y|-log|cx|$=-\frac{1}{x}$

$\log \left|\frac{y}{cx}\right|=-\frac{1}{x}$

$\frac{y}{cx}=e^{-\frac{1}{x}}$

$y=cxe^{-\frac{1}{x}}$

29.प्रारम्भिक मान प्रश्न [Solve the initial value problem] $y'=y\cot 2x , y\left(\frac{\pi}{4}\right)=2$ को हल करे।
Sol :
y'=ycot2x

$\frac{dy}{dx}=y\cot 2x$

$\frac{dy}{y}=\cot 2x dx$

Integrating both sides

$\int \frac{dy}{dx}=\int \cot 2x dx$

log|y|$=\frac{1}{2}\log |\sin 2x|+\log |c|$

2log|y|=log|c sin 2x|

log|y²|=log|c sin2x|

y²=c sin 2x

∵$y\left(\frac{\pi}{4}\right)=2$

$2^2=c \sin 2\left(\frac{\pi}{4}\right)$

4=c

∴Solution of differentiating equation

30.अवकल समीकरण [Solve the differential equation] $(x-1)\frac{dy}{dx}=2xy$ को हल करे ।
Sol :
$(x-1)\frac{dy}{dx}=2xy$

$\frac{dy}{y}=\frac{2x}{x-1}dx$

Integrating both sides

$\int \frac{dy}{y}=\int \frac{2x}{x-1}dx$

$\int \frac{dy}{y}=\int \left(2+\frac{2}{x-1}\right)dx$

log|y|=2x+2log|x-1|+c

31.अवकल समीकरण [Solve the differential equation] (1+x²)dy=xydy को हल करे।
Sol :
(1+x²)dy=xydy

$\frac{dy}{y}=\frac{x}{1+x^2}dx$

Integrating both sides

$\int \frac{dy}{y}=\frac{1}{2}\int \frac{2x}{1+x^2}dx$

log|y|$=\frac{1}{2}\log |1+x^2|+\log k$

log |y²|=log|1+x²|+log k

log |y²|=log|k(1+x²)|

y²=k(1+x²)

$y=\sqrt{k}\sqrt{1+x^2}$

y=$c\sqrt{1+x^2}$
जहाँ c=√k

32.अवकल समीकरण [Solve the differential equation] xy(y+1)dy=(x²+1)dx को हल करे।
Sol :
xy(y+1)dy=(x²+1)dx

y(y+1)dy$=\frac{(x^2+1)}{x}dx$

$(y^2+y)dy=\left(x+\frac{1}{x}\right)dx$

Integrating both sides

$\int (y^2+y)dy=\int (x+\frac{1}{x})dx$

$\frac{y^3}{3}+\frac{y^2}{2}=\frac{x^2}{2}+\log|x|+c$

33.अवकल समीकरण [Solve the differential equation] $5\frac{dy}{dx}=e^{x}y^4$ को हल करे।
Sol :
$5\frac{dy}{dx}=e^{x}y^4$

$\frac{5dy}{y^4}=e^xdx$

5y^-4 dy=e^x dx

Integrating both sides

$5\int y^{-4}dy=\int e^x dx$

$\frac{5y^{-3}}{-3}=e^x+c$

$\frac{-5}{3y^3}=e^x+c$

34.अवकल समीकरण  को हल करे। [Solve the differential equation]

sin³ xdx-sin ydy=0 

Sol :

-sin ydy=-sin³ xdx

Integrating both sides

$\int \sin ydy=\frac{1}{4}\int 4\sin ^3 xdx$

$\int \sin ydy=\frac{1}{4}\int (3\sin x-\sin 3x)dx$

$-\cos y+c=\frac{1}{4} \left[-3 \cos x+\frac{\cos 3x}{3}\right]dx$

$-\cos y+c=\frac{-3}{4}\cos x+\frac{1}{12}\cos 3x$

$c=\cos y-\frac{3}{4}\cos x+\frac{1}{12}\cos 3x$

35.निम्नलिखित अवकल समीकरणो को हल करे।

[Solve the following differential equations]:

$x\frac{dy}{dx}+y=y^2$

Sol :
$x\frac{dy}{dx}=y^2-y$

$\frac{dy}{y^2-y}=\frac{dx}{x}$

$\frac{dy}{y(y-1)}=\frac{dx}{x}$

Integrating both sides

$\int \frac{dy}{y(y-1)}=\int \frac{dy}{x}$

$\int \left[\frac{1}{y-1}-\frac{1}{y}\right]dy=\int \frac{dx}{x}$

log|y-1|-log|y|=log|x|+log c

log|y-1|=log|cxy|

y-1=cxy

36.$\frac{dy}{dx}=\frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}$

Sol :

$dy=\frac{3e^{2x}(1+e^{2x})}{e^x+\frac{1}{e^x}}dx$

$dy=\dfrac{3e^{2x}(1+e^{2x})}{\frac{e^{2x}+1}{e^x}}dx$

dy=3e^3x dx

Integrating both sides

$\int dy=3\int e^{3x}dx$

$y=3\frac{e^{3x}}{3}+c$

y=e^3x +c

37.$\sqrt{1+x^2}dy+\sqrt{1+y^2}dx=0$

Sol :

$\sqrt{1+x^2}dy=-\sqrt{1+y^2}dx=0$

$\frac{dy}{\sqrt{1-y^2}}=\frac{-dx}{\sqrt{1+x^2}}$

Integrating both sides

$\int \frac{dy}{\sqrt{1^2+y^2}}=-\int \frac{dx}{\sqrt{1^2+x^2}}$

$\log|y+\sqrt{1+y^2}|=-\log|x+\sqrt{1+x^2}|+\log c$

log|(y+√1+y²).(x+√1+x²)|=log c

(y+√1+y²)(x+√1+x²)=c

38.$\tan y\frac{dy}{dx}=\sin (x+y)+\sin (x-y)$
Sol :
$\tan y\frac{dy}{dx}=\sin x \cos y+\cos x \sin y+\sin x \cos y-\cos x \sin y$

$\tan y\frac{dy}{dx}=2\sin xdx$

sec y tan y dy=2 sin xdx

Integrating both sides

$\int \sec y \tan y dy=2\int \sin x dx$

sec y=-2cos x+c

39.अवकल समीकरण $\frac{dy}{dx}=-4xy^2$ का विशिष्ट हल ज्ञात कीजिए यदि दिया है कि y=1 जब x=0
Sol :
$\frac{dy}{dx}=-4xy^2$

$\frac{dy}{y^2}=-4xdx$

Integrating both sides

$\int \frac{dy}{y^2}=-4 \int xdx$

$-\frac{1}{y}=-4\frac{x^2}{2}+c$

$2x^2-\frac{1}{y}=c$

y=1 जब x=0

$2(0)^2-\frac{1}{1}=c$

-1=c

∴विशिष्ट हल

$2x^2-\frac{1}{y}=-1$

$2x^2+1=\frac{1}{y}$

$\frac{1}{2x^2+1}=y$

40.अवकल समीकरण $\frac{dy}{dx}=y \tan x$ का विशिष्ट हल ज्ञात कीजिए यदि दिया है कि y=1 जब x=0

Sol :

$\frac{dy}{dx}=y \tan x$

$\frac{dy}{y}=\tan x dx$

Integrating both sides

$\int \frac{dy}{y}=\int \tan x dx$

log|y|=log|sec x|+log c

log|y|=log |c sec x|

y=c sec x

y=1 , जब x=0

1=c sec 0

1=c

∴अवकल समीकरण का विशिष्ट हल 

y=sec x

41.अवकल समीकरण $x(x^2-1)\frac{dy}{dx}=1$ का विशिष्ट हल ज्ञात कीजिए यदि दिया है कि x=2 जब y=0

Sol :

$x(x^2-1)\frac{dy}{dx}=1$

$dy=\frac{dx}{x(x^2-1)}$

$dy=\left[\frac{x}{x^2-1}-\frac{1}{x}\right]dx$

Integrating both sides

$\int dy=\int \left(\frac{x}{x^2-1}-\frac{1}{x}\right)dx$

$\int dy=\frac{1}{2}\int \frac{2x}{x^2-1}dx-\int \frac{1}{x}dx$

$y=\frac{1}{2}\log|x^2-1|-\log|x|+c$

x=2 , y=0 पर

$0=\frac{1}{2}\log|2^2-1|-\log|2|+c$

$0=\frac{1}{2}\log 3-log 2+c$

$c=-\frac{1}{2}\log 3+\log 2$

∴अवकल समीकरण का हल

$y=\frac{1}{2}\log|x^2-1|-\log|x|+\log 2-\frac{1}{2}\log 3$

42.अवकल समीकरण cos ydy+cos x sinydx=0 का विशिष्ट हल ज्ञात कीजिए यदि दिया है कि $y=\frac{\pi}{2}$ जब $x=\frac{\pi}{2}$

Sol :

cos ydy+cos x sin ydx=0

cos ydy=-cos x sin ydx

$\frac{\cos y}{\sin y}dy=-\cos xdx$

Integrating both sides

$\int \frac{\cos y}{\sin y}dy=-\int cos xdx $

log|sin y|=-sin x+c

log|sin  y|+sin x=c

At $x=\frac{\pi}{2} , y=\frac{\pi}{2}$

$\log|\sin \frac{\pi}{2}|+\sin \frac{\pi}{2}=c$

log 1+1=c

1=c

∴अवकल समीकरण का हल

log|sin y|+sin x=1

43.अवकल समीकरण $(x-1)\frac{dy}{dx}=2xy$ को हल करे यदि y=1 जब x=2 

Sol :

$(x-1)\frac{dy}{dx}=2xy$

$\frac{dy}{y}=\frac{2x}{x-1}dx$

Integrating both sides

$\int \frac{dy}{y}=2\int \frac{x}{x-1}dx$

$\int \frac{dy}{y}=2 \int \left[1+\frac{1}{x-1}\right]dx$

log y=2[x+log(x-1)]+c

log y=2x+2log(x-1)+c

log y-log(x-1)²=2x+c

$\log \frac{y}{(x-1)^2}=2x+c$

y=1 जब x=2

$\log \frac{1}{(2-1)^2}=2(2)+c$

$\log \frac{1}{1}=4+c$

c=-4

∴अलकल समीकरण का हल:

$\log \frac{y}{(x-1)^2}=2x-4$

$\log \frac{y}{(x-1)^2}=2(x-2)$

$\frac{y}{(x-1)^2}=e^{2(x-2)}$

y=(x-1)²e^2(x-2)

44..अवकल समीकरण [Solve the differential equation] (1+y)²dx+xdy=0 यदि y(1)=1 [given that y(1)=1]

Sol :

(1+y)²dx+xdy=0

xdy=-(1+y²)dx

$\frac{dy}{1+y^2}=-\frac{1}{x}dx$

Integrating both sides

$\int \frac{dy}{1+y^2}=-\int \frac{1}{x}dx$

tan^-1 y-log x+c

log x+tan^-1 y=c

y=1, जब x=1

log 1+tan^-1 1=c

$\frac{\pi}{4}=c$

∴अवकल समीकरण का हल 

$\log x+\tan ^{-1}y=\frac{\pi}{4}$

45.$\cos \frac{dy}{dx}=a$ का विशिष्ट हल ज्ञात करे यदि y=2 जब x=0

Sol :

$\cos \frac{dy}{dx}=a$

$\frac{dy}{dx}=\cos ^{-1}a$

dy=cos^-1a

Integrating both sides

$\int dy=\cos ^{-1}a \int dx$

y=x cos^-1a+c

y=2 , जब x=0

2=(0)cos^-1a+c

2=c

∴अवकल समीकरण का विशिष्ट हल:

y=xcos^-1a+2

46.उस वक्र का समीकरण ज्ञात करे जो बिन्दु(1,1) से गुजरे तथा अवकल समीकरण xdy=(2x²+1)dx ,x≠0 है

Sol :

xdy=(2x²+1)dx

$dy=\frac{(2x^2+1)}{x}dx$

$dy=\left(2x+\frac{1}{x}\right)dx$

Integrating both sides

$\int dy=\int \left(2x+\frac{1}{x}\right)dx$

$y=2\frac{x^2}{2}+\log|x|+c$

y=x²+log|x|+c

x=1 ,y=1

c=0

∴वक्र का समीकरण

y=x²+log|x|

47.उस वक्र का समीकरण ज्ञात करे जो बिन्दु(0,0) से गुजरे तथा अवकल समीकरण y'=e^x sin x है

Sol :

y'=e^x sin x

$\frac{dy}{dx}=e^x \sin x$

dy=e^x sin x dx

Integrating both sides

$\int dy=\int e^x \sin x dx$

$y=\int e^x \sin x dx$

Let $I=\int e^x \sin x dx$

$I=\sin x \int e^x dx-\int \left\{\frac{d(\sin x)}{dx}\int e^x dx\right\}dx$

$I=e^x \sin x-\int \cos x e^x dx$

$I=e^x \sin x-\left[\cos x \int e^x dx-\int \left\{\frac{d(\cos x)}{dx}\int e^x dx\right\}dx\right]$

$I=e^x \sin x-\left[e^x \cos x-\int \sin x e^x dx\right]$

$I=e^x \sin x-e^x \cos x-I$

2I=e^x(sin x-cos x)

$I=\frac{1}{2}e^x \left(\sin x-\cos x\right)$

$y=\int e^x \sin xdx$

$y=\frac{1}{2}e^x (\sin x-\cos x)+c$

$y=\frac{1}{2}e^x (\sin x-\cos x)+c$

At x=0, y=0

$0=\frac{1}{2}e^0(\sin 0-\cos 0)$

$0=\frac{1}{2}(0-1)+c$

$\frac{1}{2}=c$

वक्र का समीकरण 

$y=\frac{1}{2}e^x(\sin x-\cos x)+\frac{1}{2}$

$2y=e^x(\sin x-\cos x)$

48.एक गोलाकार गुब्बारे का आयतन, जिसे हवा भरकर फुलाया जा रहा है, स्थिर गति से बदल रहा है यदि आरंभ मे इस गुब्बारे की त्रिज्या 3 इकाई है और 3 सेकेंड बाद 6 इकाई है, तो t सेकेंड बाद उस गुब्बारे की त्रिज्या ज्ञात कीजिए।

Sol :

माना गोलाकार गुब्बारे की त्रिजया r तथा आयतन=v

आयतन मे परिवर्तन की दर, $\frac{dv}{dt}=k$ (माना)

$\dfrac{d\left(\frac{4}{3}\pi r^3\right)}{dt}=k$

$4\pi r^2 \frac{dr}{dt}=k$

4πr² dr=k dt

Integrating both sides

$4\pi \int r^2 dr=k\int dt$

$\frac{4}{3}\pi r^3=kt+c$

जब, r=3, t=0

$\frac{4}{3}\pi (3)^3$

=k(0)+c

36π=c

जब, r=6, t=3

$\frac{4}{3}\pi(1)^3=k(3)+36\pi$

288π=3k+36π

252π=3k

$k=\frac{252 \pi}{3}$

k=84π

c=36π ,k=84k , t समय मे त्रिज्या

$\frac{4}{3}\pi r^3=84 \pi t+36 \pi$

r³ =63t+27

$r=(63t+27)^{\frac{1}{3}}$

49.किसी बैक मे मूलधन की वृद्धि r% वार्षिक की दर से होती है। यदि 100 रूपये 10 वर्षो मे दुगुने हो जाते है, तो r का मान ज्ञात कीजिए । (log_e=20.6931)

Sol :

माना t समय मे मूलधन P है।

$\frac{dP}{dt}=P$ का r%

$\frac{dP}{dt}=P\times \frac{r}{100}$

$\frac{dP}{dt}=\frac{r}{100}dt$

Integrating both sides

$\int \frac{dP}{p}=\frac{r}{100}\int dt$

log|P|$=\frac{r}{100}t+C$

जब ,t=0, P=100

log|100|$=\frac{r}{100}\times 0+C$

c=log 100

जब ,t=10 वर्ष , P=200

log 200$=\frac{r}{100}(10)+log 100$

log 200-log 100=$\frac{r}{10}$

$\log \left|\frac{200}{100}\right|=\frac{r}{10}$

0.6931$=\frac{r}{10}$

r=6.931%