Let xdydx−y=x2xex+ex−1∀x∈R−{0} such that y(1)=e−1. If y(2)=ky(1)(y(1)+2) the k is

dy dx−1xy=xxex+ex−1I.F=e−log⁡x=1x Solution of the differential equation in y·1x=∫ex(x+1)−1dxyx=∫xexdx+∫exdx-∫1dx⇒yx=xex-∫exdx+∫exdx−x+c⇒yx=xex−x+c ----(1) Put x=1,y=e−1e−1=e−1+c⇒c=0(1) becomes ⇒x+yx=xexx=2⇒2+y2=2e2y2=2e2−1y=4e2−4=4(e2-1) y(1)=e−1y(2)=4(e−1)(e+1)=4y(1)(y(1)+2)⇒k=4.