Mathematics - Differential Equation Question with Solution | TestHub

The given differential equation $\frac{dy}{dx}=(x-y{)}^2$ Let $x-y=t\Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}$ $\Rightarrow \frac{dy}{dx}=1-\frac{dt}{dx}$ Now, from equation (1) $\left(1-\frac{dt}{dx}\right)=(t{)}^2$ $\Rightarrow 1-t^2=\frac{dt}{dx}\Rightarrow \int dx=\int \frac{dt}{1-t^2}$ $\Rightarrow -x=\frac{1}{2\times 1}\ln \left|\frac{t-1}{t+1}\right|+c$ $\Rightarrow -x=\frac{1}{2}\ln \left|\frac{x-y-1}{x-y+1}\right|+c$ $\because$ The given condition $y(1)=1$ $-1=\frac{1}{2}\ln \left|\frac{1-1-1}{1-1+1}\right|+c\Rightarrow c=-1$ Hence, $2(x-1)=-\ln \left|\frac{1-x+y}{1-y+x}\right|$