Mathematics - Differential Equation Question with Solution | TestHub

The given differential equation can be rewritten as $\frac{d}{dx}(xy)=\frac{(\ln x{)}^2}{x}$.

Integrating both sides with respect to $x$: $xy=\int \frac{(\ln x{)}^2}{x}dx$. Let $u=\ln x$, so $du=\frac{1}{x}dx$. The integral becomes $\int u^{2}du=\frac{u^3}{3}+C$. Substituting back $u=\ln x$, we get $xy=\frac{(\ln x{)}^3}{3}+C$. Therefore, $y(x)=\frac{(\ln x{)}^3}{3x}+\frac{C}{x}$. Using the initial condition $y(e)=\frac{1}{e}$: $\frac{1}{e}=\frac{(\ln e{)}^3}{3e}+\frac{C}{e}=\frac{1}{3e}+\frac{C}{e}$. Multiplying by $e$, we get $1=\frac{1}{3}+C$, so $C=\frac{2}{3}$. Thus, the particular solution is $y(x)=\frac{(\ln x{)}^3}{3x}+\frac{2}{3x}$. Now, we find $y(e^2)$: $y(e^2)=\frac{(\ln e^2{)}^3}{3e^2}+\frac{2}{3e^2}=\frac{2^3}{3e^2}+\frac{2}{3e^2}=\frac{8}{3e^2}+\frac{2}{3e^2}=\frac{10}{3e^2}$.