Find the general solution of the equation $u_t(x,t)+t^{\frac{1}{3}}u_x(x,t)=u(x,t)$. Find the general solution of the equation $u_t(x,t)+t^{\frac{1}{3}}u_x(x,t)=u(x,t)$.
I'm having trouble starting this one.  Any help would be greatly appreciated.
 A: $$\frac{\partial u}{\partial t}+t^{1/3}\frac{\partial u}{\partial x}=u$$
Let $u(x,t)=e^{v(x,t)}$
$$\frac{\partial v}{\partial t}+t^{1/3}\frac{\partial v}{\partial x}=1$$
Let $v(x,t)=w(x,t)+t$
$$\frac{\partial w}{\partial t}+t^{1/3}\frac{\partial v}{\partial x}=0$$
$$\frac{\partial w}{t^{1/3}\partial t}+\frac{\partial v}{\partial x}=0$$
Let $T=\frac{3}{4}t^{4/3}$ then $dT=t^{1/3}dt$
$$\frac{\partial w}{\partial T}+\frac{\partial v}{\partial x}=0$$
The general solution is on the form :
$$w=f(x-T)$$
any derivable function $f$
$$w=f\left(x-\frac{3}{4}t^{4/3}\right)$$
$$v(x,t)=t+f\left(x-\frac{3}{4}t^{4/3}\right)$$
$$u(x,t)=\exp{\left(t+f\left(x-\frac{3}{4}t^{4/3}\right)\right)}$$
$$u(x,t)=e^t F\left(x-\frac{3}{4}t^{4/3}\right)$$
any derivable function $F$
A: Fourier transform on both sides on variable $x$:
$\int_{- \infty}^\infty (u_t+t^{1/3}u_x)e^{-i kx}dx = F(u_t)+t^{1/3}F(u_x)=F(u)$. Using $F(u_x)=ikF(u)+B.T.$ ($B.T.$ denotes boundary terms) one obtains the ordinary differential equation:
$\frac{dF(u)}{dt} + ikt^{1/3} F(u) = F(u)$ (here, boundary conditions are set to 0 for $x \rightarrow \infty,x \rightarrow - \infty$).
Separation of variables: $\frac{dF(u)}{F(u)} = (1-ikt^{1/3})dt$. Integration yields:
$ln|F(u)|=c+t-\frac{3}{4}ikt^{4/3}$.
Hence: $F(u) = Ce^{t-\frac{3}{4}ikt^{4/3}}$.
Finally, applying the inverse fourier transform: $u= \frac{1}{2 \pi} Ce^t \int_{- \infty}^\infty e^{ik(-3/4 t^{4/3}+x)} dk$. This integral will be a Delta Distribution.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=t^\frac{1}{3}=s^\frac{1}{3}$ , letting $x(0)=x_0$ , we have $x=x_0+\dfrac{3s^\frac{4}{3}}{4}=x_0+\dfrac{3t^\frac{4}{3}}{4}$
$\dfrac{du}{ds}=u$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^s=f\left(x-\dfrac{3t^\frac{4}{3}}{4}\right)e^t$
