\(y'+2xy=x^3\)
\(P(x)=2x\)
& \(r(x)=x^3\)
Gereduceerde formule:\(y'+2xy=0\)
\(y'=-2xy\)
\(\frac{dy}{dx}+2xy=0\)
\(\frac{dy}{y}=-2x\)
\(\int\frac{dy}{y}=\int2xdx\)
\(\ln|y|=x^2\)
\(y=e^{x^2}\)
\(\varphi\left(x\right)=\int\frac{r\left(x\right)}{p\left(x\right)}dx+C\)
\(\varphi=\int\frac{x^3}{e^{x^2}}dx\)
\(-ue^u-\frac{1}{e^u}+C\)
-x2 substituëren door u\( -x^2=u\)
\(-2xdx=du\)
\(xdx=-\frac{1}{2}du\)
\(\int\frac{x^2*x}{e^{x^2}}dx\)
Nu partieel integreren:\(dg=\frac{1}{e^u} \Rightarrow g=-\frac{1}{e^u}\)
\(f=u \Rightarrow df=du\)
\(-\frac{u}{e^u}+\int\frac{1}{e^u}du\)
\(-\frac{u}{e^u}-\frac{1}{e^u}+C\)
\(-\frac{1}{2}\left(\frac{-u-1}{e^u}\right)\)
Antwoord:\(-\frac{1}{2}\left(\frac{-x^2-1}{e^{x^2}}\right)+C\)
\(\frac{\frac{1}{2}x^2+\frac{1}{2}}{e^{x^2}}+C\)
(20 minuten verder 
)Nou zegt mijn antwoord model dat het antwoord
\(y=\frac{1}{2}x^2-\frac{1}{2}+\frac{C}{e^{x^2}}\)
moet zijn.Wat zie ik niet of doe ik fout!?!?!
