# Solution satisfying initial conditions for a pde of second order

## 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 im 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 im not actually sure ths helps.

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

Thanks

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:

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

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

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.

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