JEE Main 2019 — Mathematics Calculus

The given differential equation can be written as

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

Now, integrating factor $\displaystyle I.F.={e}^{\int \frac{1}{x}dx}={e}^{lnx}=x$I.F.=e∫x1​dx=elnx=x

Hence, the solution of the given differential equation is

$$y\cdot x=\int x\cdot {\mathrm{log}}_{e}x dx$$y⋅x=∫x⋅loge​xdx

$\displaystyle \Rightarrow y\cdot x={\mathrm{log}}_{e}x\int xdx-\int {\frac{1}{x}\int xdx}dx$⇒y⋅x=loge​x∫xdx−∫x1​∫xdxdx (using integration by parts)

$$\Rightarrow y\cdot x={\mathrm{log}}_{e}x(\frac{{x}^{2}}{2})-\int \frac{1}{x}\cdot \frac{{x}^{2}}{2}dx$$⇒y⋅x=loge​x(2x2​)−∫x1​⋅2x2​dx

$$\Rightarrow y\cdot x={\mathrm{log}}_{e}x(\frac{{x}^{2}}{2})-\int \frac{x}{2}dx$$⇒y⋅x=loge​x(2x2​)−∫2x​dx

$$\Rightarrow xy(x)=\frac{{x}^{2}}{2}{\mathrm{log}}_{e}x-\frac{{x}^{2}}{4}+C...(1)$$⇒xy(x)=2x2​loge​x−4x2​+C...(1)

Putting $x=2$x=2 we get,

$$\Rightarrow 2y(2)=2{\mathrm{log}}_{e}2-1+C$$⇒2y(2)=2loge​2−1+C

$$\Rightarrow 2y(2)={\mathrm{log}}_{e}4-1+C$$⇒2y(2)=loge​4−1+C

Given, $2y(2)={\mathrm{log}}_{e}4-1$2y(2)=loge​4−1

Then we get, $C=0$C=0

Now, from the equation $(1)$(1) we get,

$$y(x)=\frac{x}{2}{\mathrm{log}}_{e}x-\frac{x}{4}$$y(x)=2x​loge​x−4x​

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

$$y(e)=\frac{e}{2}{\mathrm{log}}_{e}e-\frac{e}{4}=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}$$y(e)=2e​loge​e−4e​=2e​−4e​=4e​