JEE Main 2019 — Differential Equations Question with Solution

The given differential equation can be written as

$\frac{dy}{dx}+\frac{y}{x}={\log }_{e}x$, which is a linear differential equation.

Now, integrating factor $I.F.=e^{\int \frac{1}{x}dx}=e^{lnx}=x$

Hence, the solution of the given differential equation is

$y·x=\int x·{\log }_{e}x dx$

$\Rightarrow y·x={\log }_{e}x\int xdx-\int \left\{\frac{1}{x}\int xdx\right\}dx$ (using integration by parts)

$\Rightarrow y·x={\log }_{e}x\left(\frac{x^2}{2}\right)-\int \frac{1}{x}·\frac{x^2}{2}dx$

$\Rightarrow y·x={\log }_{e}x\left(\frac{x^2}{2}\right)-\int \frac{x}{2}dx$

$\Rightarrow xy\left(x\right)=\frac{x^2}{2}{\log }_{e}x-\frac{x^2}{4}+C ...\left(1\right)$

Putting $x=2$ we get, 

$\Rightarrow 2y\left(2\right)=2{\log }_{e}2-1+C$

$\Rightarrow 2y\left(2\right)={\log }_{e}4-1+C$

Given, $2y\left(2\right)={\log }_{e}4-1$

Then we get, $C=0$ 

Now, from the equation $\left(1\right)$ we get, 

$y\left(x\right)=\frac{x}{2}{\log }_{e}x-\frac{x}{4}$

Putting $x=e$ in above equation, we get,

$y\left(e\right)=\frac{e}{2}{\log }_{e}e-\frac{e}{4}=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}$