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(D) Given differential equation is $\frac{dy}{dx}=(x-y)^2$ $(i)$.Let $x-y=t$. Then $1-\frac{dy}{dx}=\frac{dt}{dx}$,which implies $\frac{dy}{dx}=1-\fra

Vedclass · Accepted Answer

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

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(D) Given differential equation is $\frac{dy}{dx}=(x-y)^2$ $(i)$.Let $x-y=t$. Then $1-\frac{dy}{dx}=\frac{dt}{dx}$,which implies $\frac{dy}{dx}=1-\frac{dt}{dx}$.Substituting into $(i)$,we get $1-\frac{dt}{dx}=t^2$,so $\frac{dt}{dx}=1-t^2$.Separating variables,$dx = \frac{1}{1-t^2} dt$.Integrating both sides,$x = \int \frac{1}{1-t^2} dt = \frac{1}{2} \log \left|\frac{1+t}{1-t}\right| + c$.Substituting $t=x-y$,we get $x = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| + c$.Given $y(1)=1$,at $x=1$,$y=1$: $1 = \frac{1}{2} \log \left|\frac{1+1-1}{1-1+1}\right| + c \Rightarrow 1 = \frac{1}{2} \log(1) + c \Rightarrow c=1$.Thus,$x = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| + 1$.$x-1 = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| \Rightarrow 2(x-1) = \log \left|\frac{1+x-y}{1-x+y}\right|$.Using $\log(a/b) = -\log(b/a)$,we get $2(x-1) = -\log \left|\frac{1-x+y}{1+x-y}\right|$.

The solution of the differential equation $\frac{dy}{dx}=(x-y)^2$ when $y(1)=1$ is

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