- #1
thepatient
- 164
- 0
Given (y')^2 *y'' -8y=0 and ICs y(0)=1 y'(0)= -2
I used substitution dy/dx = v and d^2y/dx^2 = vdv/dy and got to:
v^2*v*v' = 8y
V^3 dv/dy = 8y
v^4/4 =4y + c
V^4 =4y + c
And IC y'(0) = v(0) =-2
So c = 16
v = 2(y^2+1)^(1/4)
dy/dx = 2(y^2+1)^(1/4)
(y^2+1)^(-1/4)dy = 2dx
But here I got stuck. :( did I do the steps wrong? I then tried trig substitution and got:
Sec(u)^3/2 du = 2dx
But couldn't integrate either. This is a DE class so I didn't think it should be like that and maybe I made a mistake. Any help would be greatly appreciated. :]
I used substitution dy/dx = v and d^2y/dx^2 = vdv/dy and got to:
v^2*v*v' = 8y
V^3 dv/dy = 8y
v^4/4 =4y + c
V^4 =4y + c
And IC y'(0) = v(0) =-2
So c = 16
v = 2(y^2+1)^(1/4)
dy/dx = 2(y^2+1)^(1/4)
(y^2+1)^(-1/4)dy = 2dx
But here I got stuck. :( did I do the steps wrong? I then tried trig substitution and got:
Sec(u)^3/2 du = 2dx
But couldn't integrate either. This is a DE class so I didn't think it should be like that and maybe I made a mistake. Any help would be greatly appreciated. :]