Proving an exponential function obeys the wave equation

[LiamG_G]

Homework Statement

Prove that [itex]y(x,t)=De^{-(Bx-Ct)^{2}}[/itex] obeys the wave equation

Homework Equations

The wave equation:
[itex]\frac{d^{2}y(x,t)}{dx^{2}}=\frac{1}{v^{2}}\frac{d^{2}y(x,t)}{dt^{2}}[/itex]

The Attempt at a Solution

1: [itex]y(x,t)=De^{-u^{2}}; \frac{du}{dx}=B; \frac{du}{dt}=-C[/itex]
2: [itex]\frac{dy(x,t)}{dx}=-2uBDe^{-u^{2}};

\frac{d^{2}y(x,t)}{dx^{2}}=4u^{2}B^{2}De^{-u^{2}}[/itex]
3: [itex]\frac{dy(x,t)}{dt}=2uCDe^{-u^{2}};

\frac{d^{2}y(x,t)}{dt^{2}}=4u^{2}C^{2}De^{-u^{2}}[/itex]
Then I'm stuck, I think I might have done something wrong but I can't see what.
I think v=C (from (x-vt)) and that would cancel the [itex]C^{2}[/itex] when I substitute into the wave equation, but then I would be left with a [itex]B^{2}[/itex]

 

[nasu]

The C in the exponent is not the speed of the wave. It has dimensions of s^(-1).
The speed of the wave will be a function of C and D. Which you can find from the equation.

 

[BruceW]

also, the equation
[tex]\frac{dy(x,t)}{dx}=-2uBDe^{-u^{2}}[/tex]
is correct, but the next equation
[tex]\frac{d^{2}y(x,t)}{dx^{2}}=4u^{2}B^{2}De^{-u^{2}}[/tex]
is not correct. It looks like you've done the derivative of the exponential, but what about the ##u## (in the previous equation), doesn't it depend on x also?

 

[LiamG_G]

Sorry it has taken so long. I will come back to this question I'm just in the middle of moving right now :S
Thanks for replies :)

 

[BruceW]

no worries :) hope the move goes well

 

[HallsofIvy]

Homework Statement

Prove that [itex]y(x,t)=De^{-(Bx-Ct)^{2}}[/itex] obeys the wave equation

Homework Equations

The wave equation:
[itex]\frac{d^{2}y(x,t)}{dx^{2}}=\frac{1}{v^{2}}\frac{d^{2}y(x,t)}{dt^{2}}[/itex]

The Attempt at a Solution

1: [itex]y(x,t)=De^{-u^{2}}; \frac{du}{dx}=B; \frac{du}{dt}=-C[/itex]
2: [itex]\frac{dy(x,t)}{dx}=-2uBDe^{-u^{2}};

\frac{d^{2}y(x,t)}{dx^{2}}=4u^{2}B^{2}De^{-u^{2}}[/itex]

You did not differentiate the first "u". It should be
[itex]\frac{\partial y(x,t)}{\partial x}= -2B^2De^{-u^2}- u^2B^2De^{-u^2}[/itex]

3: [itex]\frac{dy(x,t)}{dt}=2uCDe^{-u^{2}};

\frac{d^{2}y(x,t)}{dt^{2}}=4u^{2}C^{2}De^{-u^{2}}[/itex]

Then I'm stuck, I think I might have done something wrong but I can't see what.
I think v=C (from (x-vt)) and that would cancel the [itex]C^{2}[/itex] when I substitute into the wave equation, but then I would be left with a [itex]B^{2}[/itex]

 

[dextercioby]

I know that you;re supposed to differentiate like crazy to get to the expected solution, but you can surprise your professor if you make the following change of variables in your PDE: x-vt = u; x+vt = v. Then you can fully solve the original PDE and make certain simplifying assumptions to find the function you're given in the statement.