Vi bestämmer en integrerande faktor genom\[\int \sqrt{x}\, dx = \dfrac23 x^{3/2},\quad \Rightarrow \quad \textrm{I.F.}=\exp{(\dfrac{2}{3}x^{3/2})}.\]Vi multiplicerar D.E. med I.F. och får\begin{align}y'\exp{(\dfrac{2}{3}x^{3/2})}+\sqrt{x}\exp{(\dfrac{2}{3}x^{3/2})}y=\sqrt{x}\exp{(\dfrac{2}{3}x^{3/2})} \\ \Leftrightarrow\\ D\left(y\exp{(\dfrac{2}{3}x^{3/2})}\right)= \sqrt{x}\exp{(\dfrac{2}{3}x^{3/2})} \\ \Leftrightarrow \\y\exp{(\dfrac23 x^{3/2})}=\exp{(\dfrac{2}{3}x^{3/2})} +C\\ \Leftrightarrow \\ y=1+C\exp{(-\dfrac{2}{3}x^{3/2})}.\end{align}För att bestämma $C$ använder vi att $y(0)=0$:\begin{align}0=y(0)=1+Ce^0=1+C\quad \textrm{d.v.s.}\quad C=-1.\end{align}Vi får slutligen lösningen\[y=1-\exp{(-\dfrac23 x^{3/2})}.\]