Solve $2u_t-3u_x=0$ for $u(x,t)$ with initial conditions $u(x,0)=\sin x$

Question

Solve $2u_t-3u_x=0$ for $u(x,t)$ with initial conditions $u(x,0)=\sin x$

Hint: change to new coordinates

What I tried:

Assume it's separable: $$u(x,t)=X(x)T(t)$$

Substituting into the PDE gives us:

$$\frac{X'}{X}=\frac{2T'}{3T}=\lambda$$

Solving these separately gives me:

$T=e^{3/2\lambda t}$ and $X=e^{\lambda x}$. Putting these together to get:

$$u(x,t)=Ae^{3/2\lambda t}e^{\lambda x},$$which solves the equation.

Now I substitute in the initial condition which gives me

$$u(x,0)=Ae^{\lambda x}=\sin x$$

which I can't solve.

Answer 1

No, you shouldn't use here seperation of variables.

The problem is solved by characteristics:

i.e $dx/ds = -3 , dt/ds = 2 , du/ds = 0 $

So $ dx/dt = dx/ds/dt/ds = -3/2$ and we get that: $x = -3/2 t +C$ $u=f(C) = f(x+3/2 t)$. Now since $f(x) = \sin x $ we get that $u(x,t)=\sin (x+3/2 t)$.