Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& \left(1+x^2\right)\frac{dy}{dx}+xy=5x^2\sqrt{1+x^2}\\ & \frac{dy}{dx}+\frac{xy}{1+x^2}=\frac{5x^2}{\sqrt{1+x^2}}\\ & \therefore \text{ I.F. }={\mathrm{e}}^{\int \frac{\mathrm{x}}{1+x^2}\mathrm{dx}}={\mathrm{e}}^{\frac{\ln \left(1+x^2\right)}{2}}=\sqrt{1+{\mathrm{x}}^2}\\ & \therefore \mathrm{y}\sqrt{1+{\mathrm{x}}^2}=\int \frac{5{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^2}}\cdot \sqrt{1+{\mathrm{x}}^2}\mathrm{dx}\\ & \therefore \mathrm{y}\sqrt{1+{\mathrm{x}}^2}=\int \frac{5{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^2}}\cdot \sqrt{1+{\mathrm{x}}^2}\mathrm{dx}\\ & \mathrm{y}\sqrt{1+{\mathrm{x}}^2}=\frac{5{\mathrm{x}}^3}{3}+\mathrm{C}\\ & \because \mathrm{y}(0)=0\Rightarrow 0=0+\mathrm{C}\Rightarrow \mathrm{C}=0\end{array}$ $\begin{array}{rl}& \therefore y=\frac{5x^3}{3\sqrt{1+x^2}}\\ & y(\sqrt{3})=\frac{15\sqrt{3}}{32}=\frac{5\sqrt{3}}{2}\end{array}$