# Math Help - Nonlinear Equation

1. ## Nonlinear Equation

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.

2. Are you sure the substitution isn't meant to be u = y', not u = y''?

3. Yeah, I just accidentally had that messed up in the first line but I did the problem by substituting u = y'

4. (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.

5. Kudos. I have never seen a solution method like that before.

-Dan