JEE Advanced 2023 — Paper 2 — Question 1

$3 \int_{1}^{x} f(x) d t=x f(x)-\frac{x^{3}}{3}$

$\Rightarrow \quad 3 \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{xf}^{\prime}(\mathrm{x})-\mathrm{x}^{2}$ (using Newton's Leibniz theorem)

$\Rightarrow \quad \frac{d y}{d x}-\frac{2}{x} y=x$

I.F. $=\mathrm{e}^{\int-\frac{2}{x} \mathrm{dx}}=\mathrm{e}^{-2 \log \mathrm{x}}=\frac{1}{\mathrm{x}^{2}}$

$\Rightarrow$ solution is\ $\frac{y}{x^{2}}=\log _{e} x+c$

$y=x^{2} \log _{e} x+c x^{2}$ But $\mathrm{f}(1)=\frac{1}{3}$

$\Rightarrow \quad \frac{1}{3}=\mathrm{c}$

$\Rightarrow y=x^{2} \log _{\mathrm{e}} \mathrm{x}+\frac{1}{3} \mathrm{x}^{2}$

Now $y(e)=e^{2}+\frac{1}{3} e^{2}=\frac{4}{3} e^{2}$