JEE Main 2019 — Mathematics Calculus

Given, differential equation is $\frac{dy}{dx}+(\frac{2}{x})y=x$dxdy​+(x2​)y=x

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

Thus, P=\frac{2}{x}&Q=x

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

$$={e}^{\int \frac{2}{x} dx}={e}^{2\mathrm{ln}x}$$=e∫x2​dx=e2lnx

$$={e}^{\mathrm{ln}{x}^{2}}={x}^{2}.$$=elnx2=x2.

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

So, the solution of the given differential equation is

$$y{x}^{2}=\int x\cdot {x}^{2}dx+C$$yx2=∫x⋅x2dx+C

$$\Rightarrow {x}^{2}y=\int {x}^{3}dx+C$$⇒x2y=∫x3dx+C

Using $\displaystyle \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1},$∫xndx=n+1xn+1​, we get

$$\Rightarrow {x}^{2}y=\frac{{x}^{4}}{4}+C$$⇒x2y=4x4​+C

Since $y(1)=1$y(1)=1

$$\Rightarrow 1\cdot 1=\frac{1}{4}+C$$⇒1⋅1=41​+C

$$\Rightarrow C=\frac{3}{4}$$⇒C=43​

$$\Rightarrow {x}^{2}y=\frac{{x}^{4}}{4}+\frac{3}{4}$$⇒x2y=4x4​+43​

$$\Rightarrow y=\frac{{x}^{2}}{4}+\frac{3}{4{x}^{2}}.$$⇒y=4x2​+4x23​.