Question

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

Answer 1

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

Answer 2

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