Exercise 23.4
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|=ex+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|ex+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+xex)dx
dy=[sin x(sin2 x)cos2 x+xex]dx$
dy=[sin x(1-cos2x cos2 x+xex]dx$
dy=[sin x cos2 x-sin cos4 x+xex]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=ecx
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
0=x+log|1-y|+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.ex tan ydx+(1-ex)sec2 ydy=0
Sol :
(1-ex)sec2 ydy=-ex 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-ex|+log |c|
log|tan y|=log|(1-ex)c|
tan y=(1-ex)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
ey dy=(ex+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}}
e5y dy=e3x 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
3e5y -5e3x=15k
3e5y -5e3x=15k , जहाँ c=15k
Integrating both sides
\int e^{-y}dy=\int e^{x}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 sin 2x|
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
y2=4 sin 2x
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=ex 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=3e3x dx
Integrating both sides
\int dy=3\int e^{3x}dx
y=3\frac{e^{3x}}{3}+c
y=e3x +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)2e2(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
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
1=12+log|1|+c
1=1+0+c
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'=ex sin x है
y'=ex sin x
\frac{dy}{dx}=e^x \sin x
dy=ex 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=ex(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 । (loge=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%
Good
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