The solution of xdydx+2y=x^2 (x0) with y(1)=1 is:

Question

The solution of $x\dfrac{dy}{dx}+2y=x^2$ ($x
e0$) with $y(1)=1$ is:

Accepted Answer

(Same as Question 208's setup.) I.F.$=x^2$; $yx^2=\dfrac{x^4}{4}+C$; using $y(1)=1$: $C=\dfrac34$.

$$y = \frac{x^2}{4}+\frac{3}{4x^2}.$$

Solutions · 1