# Partial Differential Equations Question

1. Jun 9, 2009

### Hendrick

1. The problem statement, all variables and given/known data
Find the link between constants $$\omega$$ and $$\beta$$

so that http://www4e.wolframalpha.com/Calculate/MSP/MSP181963g2e5f4i43d3b00005ief8e24920ah323?MSPStoreType=image/gif&s=20 [Broken]
is a solution of $$\frac{\partial^{2} u}{\partial x^{2}}=2\frac{\partial u}{\partial t}$$

(A & B are constants)

2. Relevant equations
I think that $$\frac{\partial^{2} u}{\partial x^{2}}=2\frac{\partial u}{\partial t}$$ could be in the form of a 1D heat equation

3. The attempt at a solution
$$\frac{\partial^{2} u}{\partial x^{2}}=$$http://www4e.wolframalpha.com/Calculate/MSP/MSP13061963dh6ehe94f57b000031ii70cfaaf938aa?MSPStoreType=image/gif&s=24 [Broken]

$$2\frac{\partial u}{\partial t}=$$http://www4e.wolframalpha.com/Calculate/MSP/MSP12211963dh8b7ca90e2d000034i7h8i9cfgdbif7?MSPStoreType=image/gif&s=27 [Broken]

I've tried to equate the two PDEs above to solve for $$\omega$$ and $$\beta$$ but I can't work out a solution for them, therefore I think I'm going about this problem the wrong way.

Any help would be appreciated, thank you.

Last edited by a moderator: May 4, 2017
2. Jun 9, 2009

### cepheid

Staff Emeritus
What if you ignored the given solution for a while and tried solving the PDE itself in order to obtain a general solution in terms of some constants? You could then compare that general solution to the given solution in order to find omega and beta by inspection. Is this possible?

3. Jun 9, 2009

### Hendrick

Hi, do you mean try solving $$\frac{\partial^{2} u}{\partial x^{2}}=2\frac{\partial u}{\partial t}$$ without substituting u, first? Thanks

4. Jun 9, 2009

### Cyosis

I haven't looked whether you differentiated correctly, but in their current form the relation between beta and omega is pretty obvious. Factor out beta^2 in your first expression and factor out omega in your second expression, equate and divide.

5. Jun 9, 2009

### Hendrick

Hi Cyosis, I did make a mistake while equating. Thank you :)