JEE Main 2024 — Differential Equations Question with Solution

Given: $\frac{dy}{dx}=\frac{x+y-2}{x-y}$

Now, finding the intersection point of $x+y-2=0$ and $x-y=0$.

$\Rightarrow x+x-2=0, x=y$

$\Rightarrow x=1, y=1$

$\Rightarrow \left(x,y\right)\equiv \left(1,1\right)$

Let, $x=X+1, y=Y+1$

$\Rightarrow \frac{dY}{dX}=\frac{X+Y}{X-Y}$

Putting, $Y=tX$

$\Rightarrow \frac{dY}{dX}=t+X\frac{dt}{dX}$

$\Rightarrow \frac{1+t}{1-t}=t+X\frac{dt}{dX}$

$\Rightarrow \frac{1+t-t+t^2}{1-t}=X\frac{dt}{dX}$

$\Rightarrow \frac{1}{X}dX=\frac{1-t}{1+t^2}dt$

$\Rightarrow \int \frac{1}{X}dX=\int \frac{1-t}{1+t^2}dt$

$\Rightarrow \log \left|X\right|={\tan }^{-1}t-\frac{1}{2}\log \left|1+t^2\right|+C$

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

It is given that the solution curve passes through $\left(2,1\right)$.

$\Rightarrow \log \left|2-1\right|={\tan }^{-1}\left(\frac{1-1}{2-1}\right)-\frac{1}{2}\log \left|1+{\left(\frac{1-1}{2-1}\right)}^2\right|+C$

$\Rightarrow 0={\tan }^{-1}\left(0\right)-\frac{1}{2}\log \left|1\right|+C$

$\Rightarrow C=0$

$\Rightarrow \alpha =1, \beta =2$

$\Rightarrow 5\beta +\alpha =11$.