Using initial conditions in a second order PDE

[maggie56]

Homework Statement

I have a PDE for which i have found the general solution to be u(x,y) = f1(3x + t) + f2(-x + t)
where f1 and f2 are arbitrary functions. I have initial conditions u(x,0) = sin (x) and partial derivative du/dt (x,0) = cos (2x)

Homework Equations

u(x,y) = f1(3x + t) + f2(-x + t)
u(x,0) = sin (x)
du/dt (x,0) = cos (2x)

The Attempt at a Solution

I have substituted u(x,0) and du/dt (x,0) into the general solution which gives me;

u(x,0) = f1(3x) + f2(-x) = sin (x)
du/dt(x,0) = f1'(3x) + f2'(-x) = cos (2x)

but i am unsure as to where to go from here

Thanks for any help

 

[jackmell]

I'll use h and v:

h'(3x)-v'(-x)=\cos(x)
h'(3x)+v'(-x)=\cos(2x)

It's not hard to eliminate v'(-x) right?

then you'd have a regular DE:

2h'(3x)+\cos(x)+\cos(2x)

but then you got that 3x in there. What about making a change of independent variable by letting u=3x then convert the DE from one in terms of x (the one above) into one involvind u by remembering:

\frac{dh}{du}=\frac{dh}{dx}\frac{dx}{du}

now make this substitution into the one involving x, get the one involving u, solve it, then wherever there is a u in the answer, replace it by 3x.