Asked by Domuz Gribi Hayat

Solve the initial value problem 2x2y′′+xy′−3y= 0 wherey(1) = 1,y′(1) = 4.
Answered by oobleck
2x^2 y′′+xy′−3y= 0
y = c1 x^(3/2) + c2/x
now plug in the conditions