JEE Main 2019 — Differential Equations Question with Solution

Given, differential equation is $\frac{dy}{dx}+\left(\frac{2}{x}\right)y=x$

This is a linear differential equation of type $\frac{dy}{dx}+Py=Q,$ where $P\hspace{0.17em}\&\hspace{0.17em}Q$ are the functions of $x$ or constants.

Thus, $P=\frac{2}{x} \& Q=x$

The integrating factor $I.F.=e^{\int Pdx}$

$=e^{\int \frac{2}{x} dx}=e^{2\ln x}$

$=e^{\ln x^2}=x^2.$

The solution of the linear differential equation is $y\times \left(I.F.\right)=\int \left(Q\times I.F.\right)dx+C$ 

So, the solution of the given differential equation is

$y{x}^2=\int x·x^{2}dx+C$

$\Rightarrow x^{2}y=\int x^{3}dx+C$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1},$ we get

$\Rightarrow x^{2}y=\frac{x^4}{4}+C$

Since $y\left(1\right)=1$

$\Rightarrow 1·1=\frac{1}{4}+C$

$\Rightarrow C=\frac{3}{4}$

$\Rightarrow x^{2}y=\frac{x^4}{4}+\frac{3}{4}$

$\Rightarrow y=\frac{x^2}{4}+\frac{3}{4x^2}.$