We have,y+x2=dxdy⇒dxdy−y=x2It is a linear differential equation. On comparing with dxdy+Py=Q, we getP=−1,Q=x2IF=e∫Pdx=e∫−1dx=e−xRequired solution isy⋅e−x=∫x2e−xdx+c⇒ye−x=−x2e−x+2∫xe−xdx⇒ye−x=−x2e−x−2xe−x+2∫e−xdx⇒ye−x=−x2e−x−2xe−x−2e−x+c⇒y=−(x2+2x+2)+cex⇒y+x2+2x+2=cex