Find the solution of x2y''+xy'-3y=0 withy(0)=1,y'(0)=0 using laplace transform method

Question

experimentx · Answer

http://www.ijpam.eu/contents/2013-84-4/4/4.pdf

anonymous · Answer

$$\mathcal{L}\{x^2y''+xy'-3y\}=0\\frac{2}{s^3}\left(s^2Y-sy(0)-y'(0)ight)+\frac1{s^2}\left(sY-y(0)ight)-3Y=0\\frac2sY-\frac2{s^2}+\frac1sY-\frac1{s^2}-3Y=0\\left(\frac2s+\frac1s-3ight)Y=\frac2{s^2}+\frac1{s^2}\\left(\frac3s-3ight)Y=\frac3{s^2}\\left(\frac1s-1ight)Y=\frac1{s^2}\\left(s-s^2ight)Y=1\Y=-\frac1{s^2-s}=-\frac1{s(s-1)}\Y=\frac1s-\frac1{s-1}\y=\mathcal{L}^{-1}\left\{\frac1s-\frac1{s-1}ight\}=1+e^x$$