Let $y(x)$ satisfy $(x^2+1)\frac{dy}{dx}+2xy=4x^2$ with $y(0)=2$. Find $y(2)$.

Solution

1. Rewrite as $\frac{dy}{dx}+\frac{2x}{x^2+1}y=\frac{4x^2}{x^2+1}$.
2. The integrating factor is $e^{\int \frac{2x}{x^2+1}dx}=x^2+1$.
3. Multiply by the integrating factor to get $\frac{d}{dx}[(x^2+1)y]=4x^2$.
4. Integrate to get $(x^2+1)y=\frac{4}{3}{x}^3+C$.
5. Using $y(0)=2$, we find $C=2$, so $$\begin{array}{rl}y(2)& =\frac{1}{5}(\frac{32}{3}+2)\\ & =8/3\end{array}$$. Hence the answer is (B).