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$$ax^2\frac{\partial^2 v}{\partial x^2}+bx\frac{\partial v}{\partial x}+c\frac{\partial^2 v}{\partial y^2}=10x^2+9x+6$$ where $a,b,c$ are constants,

initial conditions: $v(x,0)=0,v(0,y)=0$

i tried separation method but can't get particular solution

flag	asked Jul 9 2011 at 15:56 anksh11 9●2

Separation should work. Is this homework? – András Bátkai Jul 9 2011 at 16:58

no,this is not homework – anksh11 Jul 9 2011 at 17:04

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The question to begin here would be perhaps for which values of $a$, $b$, $c\ $ a solution exists. For example, if $a=b=0\ $ then $u(x,y)=5x^2y^2/c + 3xy^2/c + 9 y^2/(2c) + C_2(x) y + C_1(x)\ $ and $u(0,y)=\frac{9 y^2}{2 c}+ C_2(0) y + C_1(0)\ $, so constants $C_1(0)$, $C_2(0)$ cannot be choosen to get $u(0,y)=0$. – Andrew Jul 10 2011 at 7:01

Robert Israel · Answer 1 · 2011-07-10 06:30:34Z UTC

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A particular solution of the pde (obtained with Maple's help) is $v(x,y) = \frac{5 x^2}{a+b} + \frac{9x}{b} + \frac{6 \ln x}{a-b}$.

answered Jul 10 2011 at 6:30

Robert Israel
10.9k●8●23

	Or, if you don't want a singularity at $x=0$, $\frac{5x^2}{a+b} + \frac{9x}{b} + \frac{3y^2}{c}$. – Robert Israel Jul 10 2011 at 16:01
	thanks man, i had tried mathematica without success but not maple ,but how you get second solution,means how we can exclude singularity while giving initial conditions/etc on maple? – anksh11 Jul 10 2011 at 18:18
	but second solution doesn't fullfill initial conditions – anksh11 Jul 10 2011 at 18:23
	Typically, when solving a nonhomogeneous boundary value problem, you take a particular solution $v_p$ of the pde (without regard to boundary conditions), and then look for a solution $v_h$ of the homogeneous pde such that $v = v_h + v_p$ satisfies the boundary conditions. Thus in your case, taking $v_p(x,y) = \frac{5x^2}{a+b} + \frac{9x}{b} + \frac{3y^2}{c}$, the boundary conditions for $v_h$ would be $v_h(x,0) = - v_p(x,0) = - \frac{5x^2}{a+b} - \frac{9x}{b}$ and $v_h(0,y) = -v_p(0,y) = - \frac{3y^2}{c}$. – Robert Israel Jul 10 2011 at 19:52
	However, you're likely to have trouble with your problem because the pde is singular at $x=0$. – Robert Israel Jul 10 2011 at 20:00

Particular solution of a non-homogenous partial differential equation.

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