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KC Sinha Solution Class 12 Chapter 23 अवकल समीकरण (Differential Equations) Exercise 23.5

 Exercise 23.5

Question 1

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

Sol :

Putting (x-y)=t

Differentiating w.r.t x

1-\frac{dy}{dx}=\frac{dt}{dx}

1-\frac{dt}{dx}=\frac{dy}{dx}

t^2 \left(1-\frac{dt}{dx}\right)=1

1-\frac{dt}{dx}=\frac{1}{t^2}

1-\frac{1}{t^2}=\frac{dt}{dx}

\frac{t^2-1}{t^2}=\frac{dt}{dx}

dx=\frac{t^2}{t^2-1}dt

Integrating both sides

\int dx=\int \frac{t^2}{t^2-1}dt

\int dx=\int \left[1+\frac{1}{t^2-1}\right]dt

x=t+\frac{1}{2\times 1}\log \left|\frac{t-1}{t+1}\right|+c

x=x-y+\frac{1}{2}\log\left|\frac{x-y-1}{x+y+1}\right|+c

y=\frac{1}{2}\log \left|\frac{x-y-1}{x-y+1}\right|+c


Question 2

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

Sol :

Putting x+y=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dy}{dx}=\frac{dt}{dx}-1

\frac{dt}{dx}-1=\cos t

\frac{dt}{dx}=1+\cos t

\frac{dt}{1+\cos t}=dx

\frac{dt}{2\cos^2 \frac{t}{2}}=dx

\frac{1}{2}\sec^2 \frac{t}{2}dt=dx

Integrating both sides

\frac{1}{2}\int \sec^2 \frac{t}{2}dt=\int dx

\dfrac{\frac{1}{2}\tan \frac{t}{2}}{\frac{1}{2}}=x+c

\tan \left(\frac{x+y}{2}\right)=x+c


Question 3

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

Sol :

Putting x+y+1=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dy}{dx}=\frac{dt}{dx}-1

t\left(\frac{dt}{dx}-1\right)=t

\frac{dt}{dx}-1=\frac{1}{t}

\frac{dt}{dx}=\frac{1}{t}+1

\frac{dt}{dx}=\frac{1+t}{t}

\frac{t}{1+t}dt=dx

\left[1-\frac{1}{1+t}\right]dt=dx

Integrating both sides

\int \left(1-\frac{1}{1+t}\right)dt=\int dx

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

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

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


Question 4

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

Sol :

Putting x+y=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dt}{dx}=e^t

\frac{dt}{e^t}=dx

e-tdt=dx

Integrating both sides

\int e^{-t}dt=\int dx

-e-t+c=x
x+e-t=c
x+e-(x+y)=c


Question 5

\frac{dy}{dx}=\sin (x+y)+\cos (x+y)

Sol :

Putting x+y=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dy}{dx}=\frac{dt}{dx}-1

\frac{dt}{dx}-1=\sin t+\cos t

\frac{dt}{dx}=\sin t+\cos t+1

\frac{dt}{\sin t+1+\cos t}=dx

\frac{dt}{2\sin \frac{t}{2}\cos \frac{t}{2}+2\cos ^2 \frac{t}{2}}=dx

\frac{dt}{2\cos \frac{t}{2}\left(\sin \frac{t}{2}+\cos \frac{t}{2}\right)}=dx

\dfrac{dt}{2\cos ^2 \frac{t}{2}\left(\frac{\sin \frac{t}{2}}{\cos \frac{t}{2}+\frac{\cos \frac{t}{2}}{\cos \frac{t}{2}}}\right)}=dx

\frac{1}{2}\dfrac{\sec ^2 \frac{t}{2}dt}{1+\tan \frac{t}{2}}=dx

Integrating both sides

\int \dfrac{\frac{1}{2}\sec^2 \frac{t}{2}dt}{1+\tan \frac{t}{2}}=\int dx

\log \left|1+\tan \frac{t}{2}\right|=x+c

log\left|1+\tan \left(\frac{x+y}{2}\right)\right|=x+c


Question 6

(x2+2xy+y2+1)\frac{dy}{dx}=2(x+y)

Sol :

Putting x+y=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dy}{dx}=\frac{dt}{dx}-1

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

\left[t^2+1\right]\left(\frac{dt}{dx}-1\right)=2t

\frac{dt}{dx}-1=\frac{2t}{t^2+1}

\frac{dt}{dx}=\frac{2t}{t^2+1}-1

\frac{dt}{dx}=\frac{2t+1+t^2}{t^2+1}

\frac{dt}{dx}=\frac{(t+1)^2}{(t+1)^2-2t}

\frac{(t+1)^2-2t}{(t+1)^2}dt=dx

\left[1-\frac{2t}{(t+1)^2}\right]dt=dx

\left[1-\frac{2t+2-2}{(t+1)^2}\right]dt=dx

\left[1-\frac{2t+2}{(t+1)^2}+\frac{2}{(t+1)^2}\right]dt=dx

Integrating both sides

\int \left[1-\frac{2t+2}{(t+1)^2}+\frac{2}{(t+1)^2}\right]dt=dx

t-\log\left|(t+1)^2\right|-\frac{2}{t+1}=x+c

t=x+c+2log|t+1|+\frac{2}{t+1}

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

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


Question 7

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

Sol :

t^2\left(1-\frac{dt}{dx}\right)=a^2

1-\frac{dt}{dx}=\frac{a^2}{t^2}

1-\frac{a^2}{t^2}=\frac{dt}{dx}

\frac{t^2-a^2}{t^2}=\frac{dt}{dx}

dx=\frac{t^2}{t^2-a^2}dt

dx=\left[1+\frac{a^2}{t^2-a^2}\right]dt

Integrating both sides

\int dx=\int \left[1+\frac{a^2}{t^2-a^2}\right]dt

x+c=t+a^2\times \frac{1}{2a}\log\left|\frac{t+a}{t-a}\right|

x+c=x-y+\frac{a}{2}\log \left|\frac{x-y+a}{x-y-a}\right|

c=\frac{a}{2}\log \left|\frac{x-y+a}{x-y-a}\right|-y


Question 8

cos(x+y)dy=dx

Sol :

Putting x+y=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dy}{dx}=\frac{dt}{dx}-1

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

\cos t\left(\frac{dt}{dx}-1\right)=1

\frac{dt}{dx}=\frac{1}{\cos t}

\frac{dt}{dx}-1=\sec t

\frac{dt}{dx}=\sec t+1

\frac{dt}{\sec t+1}=dx

\frac{dt}{\sec t+1}\times \frac{\sec t-1}{\sec t-1}=dx

\frac{\left(\sec t-1\right)}{\sec^2 t-1}dt=dx

\frac{(\sec t-1)}{\tan ^2 t}dt=dx

\left(\dfrac{\frac{1}{\cos t}}{\frac{\sin ^2 t}{\cos ^2 t}}-\frac{1}{\tan ^2 t}\right)dt=dx

\left(\frac{1}{\tan t \sin t }-\text{cot}^2 t\right)dt=dx

(cosec t cot t-cosect+1)dt=dx

Integrating both sides

\int \left(\text{cosec t } \cos t-\text{cosec}^2 t+1\right)dt=\int dx

-cosec t +cot t+t=x+c

-cosec(x+y)+cot(x+y)+x+y=x+c

-cosec(x+y)+cot(x+y)+y=c


Question 9

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

Sol :

Putting x+y=t

Differentiating w.r.t x

1+\frac{dy}{dx}=\frac{dt}{dx}

\frac{dy}{dx}=\frac{dt}{dx}-1

t^2 \left(\frac{dt}{dx}-1\right)=1

\frac{dt}{dx}-1=\frac{1}{t^2}

\frac{dt}{dx}=\frac{1}{t^2}+1

\frac{dt}{dx}=\frac{1+t^2}{t^2}

\frac{t^2}{1+t^2}dt=dx

\left[1-\frac{1}{1+t^2}\right]dt=dx

Integrating both sides

\int \left(1-\frac{1}{1+t^2}\right)dt=\int dx

t-tan-1 t=x+c

x+y-tan-1 (x+y)=x+c

y-tan-1 (x+y)=c

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