How do I find the solution to these Cauchy problems? $$ $$ 1. $$\left\{ \begin{array}{l l} yu_{x} +xu_{y}=0 & \quad \mbox{$x,y \in \mathbb{R}$}\\ u(0,y) = y^4 \end{array} \right. $$
2. $$\left\{ \begin{array}{l l} u_{x} +2u_{y}=u^2 & \quad \mbox{$x,y \in \mathbb{R}$}\\ u(x,0) = h(x) \end{array} \right. $$
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For the first problem, the textbook gives an example that the solution of the initial value problem $u_t+xu_x=0$ with $u(x,0) = f(x)$ is $u(x,t) = f(xe^{-t})$. But, I am not sure how to go about this problem since the condition is $u(0,y)=y^{4}$.
1. $$\left\{ \begin{array}{l l} yu_{x} +xu_{y}=0 & \quad \mbox{$x,y \in \mathbb{R}$}\\ u(0,y) = y^4 \end{array} \right. $$ The general solution of $yu_{x} +xu_{y}=0$ is : $$u(x,y)=F(y^2-x^2)$$ with any differentiable function $F$.
Since you didn't show what you have done, one cannot say which method you use and where exactly you got a difficulty. I will not waste time in editing a method not convenient for you.
From the wording of your question, it seems that the method of solving for the general solution is not your question, but that the difficulty arises when you have to apply the boundary condition. So, I will only edit this part of the job.
$y(0,y)=y^4=F(y^2-0)=F(y^2)$
Let $\quad y^2=X\quad\implies\quad X^2=F(X)$
The function $F$ is determined. We put it into the general solution where $\quad X=x^2-y^2\quad$. This leads to : $$u(x,y)=(y^2-x^2)^2$$
2. $$\left\{ \begin{array}{l l} u_{x} +2u_{y}=u^2 & \quad \mbox{$x,y \in \mathbb{R}$}\\ u(x,0) = h(x) \end{array} \right. $$
The general solution is : $$\frac{1}{u}+x=F(2x-y)\quad\implies\quad u(x,y)=\frac{1}{-x+F(2x-y)}$$ Same comment that above.
Boundary condition :
$u(x,0)=h(x)=\frac{1}{-x+F(2x-0)} \quad\implies\quad F(2x)=x+\frac{1}{h(x)}$
The function $F$ is determined : $\quad F(X)=\frac{X}{2}+\frac{1}{h\left(\frac{X}{2}\right)}$
We put it into the general solution where $X=2x-y$
$$u(x,y)=\frac{1}{-x+\frac{2x-y}{2}+\frac{1}{h\left(\frac{2x-y}{2}\right)}}$$
$$u(x,y)=\frac{1}{-\frac{y}{2}+\frac{1}{h\left(\frac{2x-y}{2}\right)}}$$