Help with DE

• Jul 19th 2013, 05:12 PM
Help with DE
Please could someone help me with this diff equation - I have been trying multitudes of substitution and have not yielded any progress. Help will be very appreciated.

$x^2 y' - 2yx = \dfrac{y^2}{x^2} + \dfrac{2y}{x} + x^2 + 1$

Thank you friends
• Jul 19th 2013, 06:35 PM
HallsofIvy
Re: Help with DE
That's a rather non-linear equation. There are NO general methods for solving such an equation. There are some methods for getting an approximate solution. Is that sufficient/
• Jul 20th 2013, 03:38 AM
Re: Help with DE
Nevermind
• Jul 20th 2013, 04:27 AM
Prove It
Re: Help with DE
Quote:

Please could someone help me with this diff equation - I have been trying multitudes of substitution and have not yielded any progress. Help will be very appreciated.

$x^2 y' - 2yx = \dfrac{y^2}{x^2} + \dfrac{2y}{x} + x^2 + 1$

Thank you friends

Maybe a substitution of the form \displaystyle \begin{align*} v = \frac{y}{x} \end{align*} would work.

Notice that if \displaystyle \begin{align*} v = \frac{y}{x} \end{align*}, then \displaystyle \begin{align*} y = v\,x \end{align*} and \displaystyle \begin{align*} \frac{dy}{dx} = v + x\,\frac{dv}{dx} \end{align*}, so substituting into the DE we have

\displaystyle \begin{align*} x^2\,\frac{dy}{dx} - 2\,x\,y &= \frac{y^2}{x^2} + \frac{2\,y}{x} + x^2 + 1 \\ x^2 \left( v + x\,\frac{dv}{dx} \right) - 2\,x\left( v\,x \right) &= v^2 + 2v + x^2 + 1 \\ x^2\,v + x^3\,\frac{dv}{dx} - 2\,x^2 \, v &= v^2 + 2\,v + x^2 + 1 \\ x^3\,\frac{dv}{dx} - x^2\,v &= v^2 + 2\,v + x^2 + 1 \\ \frac{dv}{dx} - x^{-1}\,v &= x^{-3}\,v^2 + 2\,x^{-3}\,v + x^{-1} + x^{-3} \\ \frac{dv}{dx} + \left( -x^{-1} - 2\,x^{-3} \right) v &= x^{-3}\,v^2 + x^{-1} + x^{-3} \end{align*}

This is ALMOST a Bernoulli DE, I wonder if anyone knows how to proceed from here...
• Jul 20th 2013, 08:41 AM
JJacquelin
Re: Help with DE
Quote:

Please could someone help me with this diff equation - I have been trying multitudes of substitution and have not yielded any progress. Help will be very appreciated.
$x^2 y' - 2yx = \dfrac{y^2}{x^2} + \dfrac{2y}{x} + x^2 + 1$
Thank you friends

Hi !
This is a Riccati equation.
write it on the form : y' = P(x) +Q(x)*y +R(x)*y²
Let y = -z'/(z*R)
Compute y' in terms of z'', z', z
Bring back y' and y into y' = P(x) +Q(x)*y +R(x)*y²
This leads to a linear ODE (second order)
Solve it for z(x)
Compute z'
Bring back z' and z into y = -z'/(z*R) which leads to y(x)