The solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^2$ when $y(1)=1$ is
- $\quad \log \left|\frac{2-y}{2-x}\right|=2(y-1)$
- $\quad-\log \left|\frac{1+x-y}{1-x+y}\right|=x+y-2$
- $\quad \log \left|\frac{2-x}{2-y}\right|=x-y$
- $\quad-\log \left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$
Solution
Let $x-y=\mathrm{t}$ $\begin{aligned} \therefore \quad & 1-\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{dt}}{\mathrm{~d} x} \\ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=1-\frac{\mathrm{dt}}{\mathrm{~d} x} \end{aligned}$
From (i) $\begin{aligned} & 1-\frac{\mathrm{dt}}{\mathrm{~d} x}=\mathrm{t}^2 \\ & 1-\mathrm{t}^2=\frac{\mathrm{dt}}{\mathrm{dx}} \end{aligned}$ $\therefore \quad \mathrm{d} x=\frac{1}{1-\mathrm{t}^2} \mathrm{dt}$ Integrating on both sides, we get $\begin{aligned} & \int \mathrm{d} x=\int \frac{1}{1-\mathrm{t}^2} \mathrm{dt} \\ & x=\frac{1}{2} \log \left|\frac{1+\mathrm{t}}{1-\mathrm{t}}\right|+\mathrm{c} \\ & x=\frac{1}{2} \log \left|\frac{1+x-y}{1-(x-y)}\right|+\mathrm{c} \\ & x=\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right|+\mathrm{c} \end{aligned}$ $\text { But } y(1)=1$ $\begin{aligned} & \therefore \quad c=1 \\ & x=\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right|+1 \\ & x-1=\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| \\ & 2(x-1)=\log \left|\frac{1+x-y}{1-x+y}\right| \\ & \Rightarrow-\log \left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 1)