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
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-cosec2 t+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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