$$2u_t + x(1+t)u_x = u^2$$ $$u(x,0)=x$$
Tried to do this by method of characteristics.
So, using change of variables as follows $$\left\lbrace \begin{array}{c} p=x \\ q = 2\ln(x)-(t+t^2/2) \end{array}\right.$$
But I am stuck from this point onward. I seem to think, I might have done something wrong here.
Any hints? Thanks!
$$2\frac{\partial u}{\partial t}+x(1+t)\frac{\partial u}{\partial x}=u^2\quad\text{with}\quad u(x,0)=x$$We will solve the above inhomogeneous Cauchy problem with variable coefficients using the method of characteristic curves. First, note this corresponds to $\dfrac{du}{ds}$$$\frac{du}{ds}=\frac{\partial u}{\partial t}\frac{dt}{ds}+\frac{\partial u}{\partial x}\frac{dx}{ds}$$where clearly $\dfrac{du}{ds}=u^2,\dfrac{dt}{ds}=2,\dfrac{dx}{ds}=x(1+t)$. This yields $t(s)=2s$ and so $$ \frac{dx}{ds}=x(1+2s) \\ \frac{1}{x}\frac{dx}{ds}=1+2s \\ \int\frac1x\frac{dx}{ds}=\int(1+2s)~ds\\\log x=s+s^2\\ x(s)=\exp(s+s^2) $$Similarly,$$\frac{du}{ds}=u^2\\ \frac1{u^2}\frac{du}{ds}=1\\ \int\frac1{u^2}\frac{du}{ds}ds=\int ds \\ -\frac1u=s\\ u(s)=-\frac1{s}$$Can you take it from here?
The Lagrange-Charpit equations for the PDE $$ 2u_t+x(1+t)u_x=u^2 \tag{1} $$ are given by $$ \frac{dt}{2}=\frac{dx}{x(1+t)}=\frac{du}{u^2}. \tag{2} $$ Solving $(2)$ we obtain $$ \frac{dt}{2}=\frac{dx}{x(1+t)}\implies (1+t)dt=\frac{2dx}{x} \implies t+\frac{t^2}{2}-\ln x^2=C_1, \tag{3} $$ $$ \frac{dt}{2}=\frac{du}{u^2} \implies \frac{1}{u}+\frac{t}{2}=C_2. \tag{4} $$ Combining $(3)$ and $(4)$ we can write the general solution to $(1)$ as $$ \frac{1}{u}=-\frac{t}{2}+f\!\left(t+\frac{t^2}{2}-\ln x^2\right). \tag{5} $$ The function $f$ is fixed by the initial condition $u(x,0)=x$: $$ \frac{1}{x}=f(-\ln x^2). \tag{6} $$ Let $-\ln x^2=\xi$; then $x=e^{-\xi/2}$, and Eq. $(6)$ implies $f(\xi)=e^{\xi/2}$. Therefore, we can rewrite $(5)$ as $$ \frac{1}{u}=-\frac{t}{2}+\exp\left(\frac{t}{2}+\frac{t^2}{4}-\frac{1}{2}\ln x^2\right)=-\frac{t}{2}+\frac{1}{x}\exp\left(\frac{t}{2}+\frac{t^2}{4}\right). \quad{\square} \tag{7} $$
