ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Fri, 08 May 2015 16:05:55 +0200solve differential equationhttps://ask.sagemath.org/question/24486/solve-differential-equation/I need to solve this third-order linear partial differential equation:
d^2/dx^2 d/dy f(x,y) = f(x,y) - x*y
Could you please help me to do this?
Thank you very much for your advise!Wed, 15 Oct 2014 00:10:24 +0200https://ask.sagemath.org/question/24486/solve-differential-equation/Comment by kcrisman for <p>I need to solve this third-order linear partial differential equation:</p>
<p>d^2/dx^2 d/dy f(x,y) = f(x,y) - x*y
Could you please help me to do this?</p>
<p>Thank you very much for your advise!</p>
https://ask.sagemath.org/question/24486/solve-differential-equation/?comment=24500#post-id-24500Is this something that even has a symbolic solution? There is some numeric stuff that should be usable with PDEs, but I'm not as familiar with that.Thu, 16 Oct 2014 15:56:09 +0200https://ask.sagemath.org/question/24486/solve-differential-equation/?comment=24500#post-id-24500Comment by vdelecroix for <p>I need to solve this third-order linear partial differential equation:</p>
<p>d^2/dx^2 d/dy f(x,y) = f(x,y) - x*y
Could you please help me to do this?</p>
<p>Thank you very much for your advise!</p>
https://ask.sagemath.org/question/24486/solve-differential-equation/?comment=24513#post-id-24513True. It really depends on what you mean by solve... it can be a symbolic solution, an explicit (convergent) Taylor expansion at the origin, a numerical approximation, etc.Fri, 17 Oct 2014 17:31:34 +0200https://ask.sagemath.org/question/24486/solve-differential-equation/?comment=24513#post-id-24513Answer by GM3D for <p>I need to solve this third-order linear partial differential equation:</p>
<p>d^2/dx^2 d/dy f(x,y) = f(x,y) - x*y
Could you please help me to do this?</p>
<p>Thank you very much for your advise!</p>
https://ask.sagemath.org/question/24486/solve-differential-equation/?answer=26760#post-id-26760Not sure if you still need an answer, but here goes;
Define g(x) = f(x) - x*y, this makes the given equation homogeneous;
d^2/dx^2 d/dy g(x, y) = g(x, y)
And now assume g(x, y) = A(x)B(y). That lets you factorize the equation into two parts;
d^2/xx^2 A(x) = a A(x), d/dy B(y) = (1/a) B(y), with some positive constant a.
This can be solved as
A(x) = C1 exp(+-sqrt(a) x), B(y) = C2 exp(y / a)
And by multiplying them back,
g(x, y) = (C1 * C2) exp(+-sqrt(a) x + y / a) = C exp(+-sqrt(a) x + y / a)
Therefore, f(x, y) = C exp(+-sqrt(a) x + y / a) + x*y.
General solution is linear combination of these solution with different values of C and a.
But I might have missed something there, because this solution contains only two arbitrary parameters whereas the original equation is a third order differential equation.
Fri, 08 May 2015 16:05:55 +0200https://ask.sagemath.org/question/24486/solve-differential-equation/?answer=26760#post-id-26760