JEE Main 2023 — Mathematics Calculus

Given,

$$x\mathrm{ln}x\frac{dy}{dx}+y={x}^{2}\mathrm{ln}x$$xlnxdxdy​+y=x2lnx

$$\Rightarrow \frac{dy}{dx}+\frac{y}{x\mathrm{ln}x}=x$$⇒dxdy​+xlnxy​=x

This is a linear differential equation.

Now, finding integrating factor we get

$$I.F={e}^{\int \frac{1}{\mathrm{xlnx}}\mathrm{dx}}={e}^{\mathrm{ln}(\mathrm{lnx})}=\mathrm{lnx}$$I.F=e∫xlnx1​dx=eln(lnx)=lnx

Now, solution of the differential equation is given by

$$y\times I.F=\int (x\times I.F)dx$$y×I.F=∫(x×I.F)dx

$$\Rightarrow y\times \mathrm{ln}x=\int \mathrm{xlnx}dx$$⇒y×lnx=∫xlnxdx

$$\Rightarrow y\times \mathrm{ln}x=\mathrm{ln}x\cdot \frac{{x}^{2}}{2}-\frac{{x}^{2}}{4}+c$$⇒y×lnx=lnx⋅2x2​−4x2​+c

Now given $y(2)=2$y(2)=2,

$$2\times \mathrm{ln}2=\mathrm{ln}2\cdot \frac{{2}^{2}}{2}-\frac{{2}^{2}}{4}+c$$2×ln2=ln2⋅222​−422​+c

$$\Rightarrow c=1$$⇒c=1

So, equation will be $y\times \mathrm{ln}x=\mathrm{ln}x\cdot \frac{{x}^{2}}{2}-\frac{{x}^{2}}{4}+1$y×lnx=lnx⋅2x2​−4x2​+1

Now finding $y(e)$y(e) we get,

$$y\times \mathrm{ln}e=\mathrm{ln}e\cdot \frac{{e}^{2}}{2}-\frac{{e}^{2}}{4}+1$$y×lne=lne⋅2e2​−4e2​+1

$$\Rightarrow y\times 1=1\cdot \frac{{e}^{2}}{2}-\frac{{e}^{2}}{4}+1$$⇒y×1=1⋅2e2​−4e2​+1

$$\Rightarrow y=\frac{4+{e}^{2}}{4}$$⇒y=44+e2​