Solving a Second Order Non-Linear PDE with Undetermined Coefficients

[binbagsss]

Homework Statement

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

Homework Equations

This is second order non-linear pde of the 'form' $f(y'',y',y,x)$ .

I have read that there are 2 simplified versions of a second order non-linear pde that can be solved easily and these are 1) when there is no y term 2) when there is no x term

The Attempt at a Solution

[/B]
The above does not fit into these two categories, and as such I think it has no general sort of procedure or way to solve, but it is possible for an integrating factor to come for mind for this particular one? (i just have no idea what this could be !)

Many thanks

 

[tnich]

Homework Statement

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

Homework Equations

This is second order non-linear pde of the 'form' $f(y'',y',y,x)$ .

I have read that there are 2 simplified versions of a second order non-linear pde that can be solved easily and these are 1) when there is no y term 2) when there is no x term

The Attempt at a Solution

[/B]
The above does not fit into these two categories, and as such I think it has no general sort of procedure or way to solve, but it is possible for an integrating factor to come for mind for this particular one? (i just have no idea what this could be !)

Many thanks

Putting the equation in this form
$\frac{y''}{yy'}=2x$
suggests that $y$ is a polynomial or a rational function. Experiment with these forms using undetermined coefficients and you will quickly find a solution.

 

[Ray Vickson]

Putting the equation in this form
$\frac{y''}{yy'}=2x$
suggests that $y$ is a polynomial or a rational function. Experiment with these forms using undetermined coefficients and you will quickly find a solution.

This type of suggestion would not work for the simpler similar case $\frac{y'}{y} = 2x$.

Anyway, if $y$ is a polynomial of degree $n$, the right-hand-side $2 x y y'$ is a polynomial of degree $2n$ while the left-hand-side $y''$ is a polynomial of degee $n-2$. That requires $2n = n-2$, hence $n = -2$. That suggests a solution of the form $1/x^2$!