Let $y=y(x)$ be the solution of the differential equation $x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=x \log x,(x\gt1)$ If $2(y(2))=\log 4-1$ then the value of $y(\mathrm{e})$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=x \log x \\ \therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}+\frac{1}{x} y=\log x \end{aligned}$
Here, $\mathrm{P}(x)=\frac{1}{x}, \mathrm{Q}(x)=\log x$ $\begin{array}{ll} \therefore & \text { I.F. }=\mathrm{e}^{\int \mathrm{P}(x) \mathrm{dx}}=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}=\mathrm{e}^{\log x}=x \\ \therefore & y(\text { I.F. })=\int \mathrm{Q}(\text { I.F. }) \mathrm{d} x+\mathrm{c} \\ \therefore & x y=\int x \log x \mathrm{~d} x+\mathrm{c} \\ \therefore & x y=\frac{x^2}{2} \log x-\left(\int \frac{x^2}{2} \times \frac{1}{x} \mathrm{~d} x\right)+\mathrm{c} \\ \therefore & x y=\frac{x^2}{2} \log x-\frac{1}{2} \int x \mathrm{~d} x+\mathrm{c} \\ \therefore & x y=\frac{x^2}{2} \log x-\frac{1}{4} x^2+\mathrm{c}...(i) \end{array}$
Given that $2(y(2))=\log 4-1$ $\therefore \quad y(2)=\log 2-\frac{1}{2}$ $\therefore \quad$ From equation (i), we get $\begin{array}{ll} & 2\left(\log 2-\frac{1}{2}\right)=\frac{(2)^2}{2} \log 2-\frac{1}{4}(2)^2+\mathrm{c} \\ \therefore & 2 \log 2-1=2 \log 2-1+c \\ \therefore \quad & \mathrm{c}=0...(ii) \end{array}$ $\therefore \quad$ From (i) and (ii), we get $y(\mathrm{e})=\frac{\mathrm{e}}{2}-\frac{\mathrm{e}}{4}=\frac{\mathrm{e}}{4}$

Let $y=y(x)$ be the solution of the differential equation $x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=x \log x…

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