Solution satisfying initial conditions for a pde of second order

[maggie56]

Homework Statement

I have found the general solution to a second order pde to be
U(x,t) = f(3x + t) + g(-x + t) where f and g are arbitrary functions

I have initial conditions
U(x,0) = sin(x)
Du/dt (x,0) = cos (2x)

The Attempt at a Solution

I have found that
U(x,0) = f(3x) + g(-x) = sin(x)
Du/dt(x,0) = f'(3x) + g'(-x) = cos(2x)

From this point I am not sure what to do. I have tried differentiating u(x,0) with respect to x which i think gives me
Du/dx (x,0) = 3f'(3x) - g'(-x) = cos (x)
I thought i would then equate the equations but I am not actually sure ths helps.

Could someone please point me in the right direction for this question

Thanks

 

[jackmell]

Ok, didn't get it that other time huh? Then I'll work one through and you do the other one ok.

$$f(3x)+g(-x)=\sin(x)$$
$$f'(3x)+g'(-x)=\cos(x)$$

I differentiate the top one with respect to x:

$$3f'(3x)-g'(-x)=\cos(x)$$
$$f'(3x)+g'(-x)=\cos(x)$$

Now add them:

$$4f'(3x)=2\cos(x)$$

To get rid of that 3x in there, let u=3x or x=1/3u then:

$$\frac{df}{du}=\frac{df}{dx}\frac{dx}{du}=1/3\frac{df}{dx}$$

Then substitute that into that DE with 3x in there:

$$12 \frac{df}{du}=2\cos(1/3u)$$

integrate it:

$$f(u)=1/2\sin(u/3)$$

so that:

$$f(3x+t)=1/2 \sin((3x+t)/3)$$

I believe that's right anyway. Would need to back-substitute it into the PDE to check it.

Ok, now you do the other one.