Problem: Solve the initial Value: when x=1, y=0 dy/dx = 1 2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0 My attempt: x = e^t dx/dt = e ^t dy/dt = dy/dx * dx/dt dy/dt = x*dy/dx d^2y/dt^2 = d/dt(dy/dt) = d/dt(x*dy/dx) =d/dx(x*dy/dx)*(dx/dt) since dx/dt = x =(x^2*d^2y/dx^2) + (x*dy/dx) =(x^2*d^2y/dx^2) + (dy/dt) (x^2*d^2y/dx^2) = (d^2y/dt^2) - (dy/dt) From this point, do I simply substitute the Values for the 1st/2nd derivatives in ? i.e. 2((d^2y/dt^2) - (dy/dt)) + 3(dy/dt) - 15y = 0 2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0 2m^2 +3m - 1 = 0 Then solve for m, Then create the solution in the form of : Ae^mt + Be^mt = 0 ?? Is that the right path? Any help is appreciated. Cheers.