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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