Are you sure the substitution isn't meant to be u = y', not u = y''?
Solve the given DE by using the substitution u = y'
(y + 1)y" = (y')^2
(y + 1)u' = u^2
u' = u*du/dy
(y + 1)u*du/dy = u^2
(y + 1)du/dy = u
1/(y + 1) dy = (1/u) du
ln(y + 1) = ln(u) + C
y + 1 = Cu
u = (y + 1)/C = dy/dx
dx = C/(y + 1) dy
x + C2 = C*ln(y + 1)
I'd appreciate it if someone could look over this and tell me if I did it right, thanks in advance.
(y + 1)(d^2y/dx^2) = (dy/dx)^2
(y + 1)(d^2y/dx^2) = (dy/dx)(dy/dx)
After the substitution you will have
(y + 1)(du/dx) = u(dy/dx)
(1/u)(du/dx) = [1/(y+1)](dy/dx)
Now integrating both sides with respect to x gives
int[(1/u)(du/dx)dx] = int{[1/(y+1)](dy/dx)dx}
int[(1/u)du] = int{[1/(y+1)]dy}
ln|u| + C_1 = ln|y + 1| + C_2
ln|u| - ln|y + 1| = C where C = C_2 - C_1
ln|u/(y + 1)| = C
ln|(dy/dx)/(y + 1)| = C
[1/(y + 1)](dy/dx) = A where A = (+-)e^C
Integrating both sides with respect to x gives
int{[1/(y+1)](dy/dx)dx} = int(A dx)
int{[1/(y + 1)]dy} = Ax + C_3
ln|y + 1| + C_4 = Ax + C_3
ln|y + 1| = Ax + B where B = C_3 - C_4
|y + 1| = e^(Ax + B)
|y + 1| = (e^B)e^(Ax)
y + 1 = De^(Ax) where D = (+-)e^B
y = De^(Ax) - 1.