# Nonlinear ODE

• Jan 2nd 2011, 06:40 AM
Heirot
Nonlinear ODE
Being a physicist, I rarely encountered a nonlinear ODE. Accordingly, I have no idea how to solve the following equation:

y(x) * y''(x) + y'(x)^2 = 0

with y(0) = C, y'(x) = 0.

It's fairly obvious that one of the solutions is y(x) = C. But, how to find the other solution? I don't even know enough about general ODE to know is there another solution being the equation is not linear?

Thanks

*** I now see I should've posted this under the Differential Equations. Sorry! ***
• Jan 2nd 2011, 07:00 AM
Prove It
$\displaystyle \displaystyle y\,\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$

$\displaystyle \displaystyle y\,\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^2$

$\displaystyle \displaystyle \frac{d^2y}{dx^2} = -\frac{\left(\frac{dy}{dx}\right)^2}{y}$

$\displaystyle \displaystyle \frac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}} = -\frac{\frac{dy}{dx}}{y}$

Now let $\displaystyle \displaystyle u = \frac{dy}{dx}$ so that $\displaystyle \displaystyle \frac{du}{dx} = \frac{d^2y}{dx^2}$ so that

$\displaystyle \displaystyle \frac{1}{u}\,\frac{du}{dx} = -\frac{1}{y}\,\frac{dy}{dx}$

$\displaystyle \displaystyle \int{\frac{1}{u}\,\frac{du}{dx}\,dx} = \int{-\frac{1}{y}\,\frac{dy}{dx}\,dx}$

$\displaystyle \displaystyle \int{\frac{1}{u}\,du} = \int{-\frac{1}{y}\,dy}$

$\displaystyle \displaystyle \ln{|u|} + C_1 = -\ln{|y|} + C_2$

$\displaystyle \displaystyle \ln{|u|} + \ln{|y|} = C$ where $\displaystyle \displaystyle C = C_2 - C_1$

$\displaystyle \displaystyle \ln{|u\,y|} = C$

$\displaystyle \displaystyle \ln{\left|y\,\frac{dy}{dx}\right|} = C$

$\displaystyle \displaystyle \left|y\,\frac{dy}{dx}\right| = e^C$

$\displaystyle \displaystyle y\,\frac{dy}{dx} = A$ where $\displaystyle \displaystyle A = \pm e^C$

$\displaystyle \displaystyle \int{y\,\frac{dy}{dx}\,dx} = \int{A\,dx}$

$\displaystyle \displaystyle \int{y\,dy} = Ax + C_3$

$\displaystyle \displaystyle \frac{y^2}{2} + C_4 = Ax + C_3$

$\displaystyle \displaystyle \frac{y^2}{2} = Ax + B$ where $\displaystyle \displaystyle B = C_3 - C_4$

$\displaystyle \displaystyle y^2 = 2(Ax + B)$

$\displaystyle \displaystyle y = \pm \sqrt{2(Ax + B)}$.
• Jan 2nd 2011, 07:36 AM
Jester
Might be a little easier to recognize that

$\displaystyle y y'' + y'^2 = \left(y y'\right)'$.
• Jan 2nd 2011, 07:51 AM
FernandoRevilla
Another way: the function

$\displaystyle F(y,y',y'')=yy''+(y')^2$

satisfies:

$\displaystyle F(\lambda y,\lambda y',\lambda y'')=\lambda^2 F(y,y',y'')$

that is, it is homogeneous. Using the substitution:

$\displaystyle y=e^{\int z\;dx},\quad (z=z(x))$

the new equation is a first order equation.

Fernando Revilla