How do you solve Cauchy problem to the first order PDE

Question

How do you solve Cauchy problem to the first order PDE of the form: $y^{-1}u_x+u_y=u^2, u(x,1)=x^2$

Answer 1

$$y^{-1}u_x+u_y=u^2$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{y^{-1}}=\frac{dy}{1}=\frac{du}{u^2}$$ A first characteristic equation comes from solving $\frac{dx}{y^{-1}}=\frac{dy}{1}$ : $$ye^{-x}=c_1$$ A second characteristic equation comes from solving $\frac{dy}{1}=\frac{du}{u^2}$ : $$y+\frac{1}{u}=c_2$$ The general solution of the PDE on the form of implicit equation $c_2=F(c_1)$ is : $$y+\frac{1}{u}=F\big(ye^{-x}\big)$$ $F$ is an arbitrary function (to be determined according to some boundary condition).

$$\boxed{u(x,y)=\frac{1}{-y+F\big(ye^{-x}\big)}}$$

With condition $u(x,1)=x^2$ : $$x^2=\frac{1}{-1+F\big(e^{-x}\big)} \quad\implies\quad F\big(e^{-x}\big)=1+\frac{1}{x^2} $$ Let $X=e^{-x}\quad\implies\quad x=-\ln|X|$ $$F(X)=1+\frac{1}{(\ln(X))^2}$$ The function $F(X)$ is determined. We put it into the above general solution where $X=ye^{-x}$ : $$u(x,y)=\frac{1}{-y+1+\frac{1}{\big(\ln(ye^{-x})\big)^2}}$$ $$u(x,y)=\frac{1}{-y+1+\frac{1}{\big(\ln(y)-x\big)^2}}$$ This is the solution which satisfies both the PDE and the boundary condition.