# Engine on the net that will solve PDEs

1. Nov 25, 2008

### coverband

Hi

Does anybody know if there is an engine on the net that will solve PDEs.

I'm looking for the solution of a linear first order PDE

Thanks

2. Nov 25, 2008

### jambaugh

Re: Pde

Could you post the form of the PDE?

3. Nov 25, 2008

### coverband

Re: Pde

yzx+xzy+z=y z=2 @ (3,1)

4. Nov 25, 2008

### matematikawan

Re: Pde

Is it not that we first solve the auxiliary equation ?
$$\frac{dx}{y} = \frac{dy}{x} = \frac{dz}{y - z}$$

Let the two solutions be u=c1 and v=c2. Then the general solution is u=f(v).

But not sure how to include the given initial condition z=2 @ (3,1) .

5. Nov 25, 2008

### coverband

Re: Pde

Thank you for your input. I will consider this

6. Nov 25, 2008

### matematikawan

Re: Pde

I had sent the following to convode@riemann.physmath.fundp.ac.be

NMAX:=3;
SMAX:=4;
LPARTI:={};
ARGSTOP:=2;
DEPEND Z,X;
DEPEND Z,Y;
L2:={Z};
L3:={X,Y};
L4:={2,{3,1}};
L5:={};
L1:={Y*DF(Z,X)+X*DF(Z,Y)+Z-Y=0};
CONVODE(L1,L2,L3,L4,L5);

and received the reply for the solution as follows (if I retype it correctly)
$$z = (2x+2y)^{-1} (2\sqrt{- x^2 +y^2}f(- x^2 + y^2)+(y^2-x^2)\log(\frac{x+y}{\sqrt{-x^2 + y^2}})+x^2 +xy)$$

where f is an arbitrary function.
Can we believe the solution given?

7. Nov 26, 2008

### coverband

Re: Pde

Thanks

Convode sends it back in a weird asss format in the email. Difficult to read no !?

I think the question might be faulty or at least the initial conditions. I don't know is it solvable by just being given the value of z at a POINT rather than along a curve/line

Last edited: Nov 26, 2008
8. Nov 26, 2008

### matematikawan

Re: Pde

Never mind about that messy result given by convode. But it do gave something.

Let say there is a mistake giving a point. A curve or a line given instead as you said. How do we determine that arbitrary function?