# Solution satisfying initial conditions for a pde of second order

• maggie56
In summary, the conversation is about finding the general solution to a second order partial differential equation with initial conditions. The general solution is given by U(x,t) = f(3x + t) + g(-x + t) where f and g are arbitrary functions. The initial conditions are U(x,0) = sin(x) and Du/dt (x,0) = cos (2x). To solve the equation, the top equation is differentiated with respect to x and then added to the bottom equation. This results in 4f'(3x) = 2cos(x). To eliminate the 3x, u is substituted for 3x and the equation is integrated to get f(u) = 1/

## 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

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.

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