Solve pde problem

Question

For each of the following PDE. (a) solve the characteristic equation (b) define a transformation of the PDE. And obtain the transformed equation. (c) find the general solution of the transformed equation.

$$xu_x-yu_y+u =x $$ Let $r=r(r,s) ,s=s (r,s) $ Char. eq. : $$\frac{dy}{dx}= \frac {-y}{x} $$ $$\frac{dy}{y}= \frac {-dx}{x} $$ $$lny =-lnx + c$$ $$s=c =lnx+lny=lnxy $$ let r=x $$u_x=u_r +\frac {u_s}{x}$$ $$u_y=0 +\frac {u_s}{y}$$

Substitute in eq.$(r=x,s=lnxy)$ $$ru_r+u_s -u_s+u =r $$ $$ru_r+u=r $$ First one homogeneous eq $$u_r+\frac {u}{r}=0$$ $$M= exp ( \int (1/r))=r$$

$$ru_r+u=0$$ $$u = \frac {F (s)}{r}=\frac{F (lnxy) }{x}$$

How to find nonhomogenes for this problem? Thanks

Answer 1

$$ru_r+u=r$$ You want the solution for this equation ? $$r\frac {du}{dr}+u=r$$ $$(ru)'=r$$ Just integrate $$ru=\int rdr=\frac {r^2}2+K(s)$$ $$u=\frac 1r( \frac {r^2}2+K(s))$$ $$\boxed{u(r,s)=\frac {K(s)}r+ \frac {r}2}$$

Answer 2

$$x u_x-y u_y=x-u$$ $$\frac{dx}{x}=\frac{dy}{-y}=\frac{du}{x-u}$$ A first family of characteristic curves comes from $\quad \frac{dx}{x}=\frac{dy}{-y}$ $$xy=c_1$$ A second family of characteristic curves comes from $\frac{dx}{x}=\frac{du}{x-u}=\frac{-dx+2du}{-x+2(x-u)}=\frac{d(-x+2u)}{-(-x+2u)}$ $$x(-x+2u)=c_2$$ General solution of the PDE : $$x(-x+2u)=\Phi(xy)$$ $$u(x,y)=\frac12\left(x+\frac{1}{x}\Phi(xy) \right)$$ $\Phi$ is an arbitrary function, to be determined according to the boundary conditions.