We have the following differential equation: $$u_xx-2u_xy+u_yy+9u_x+9u_y-9u=0$$ The goal is to solve with the method of characteristics, in other words to simplify, to give the canonical form. But I am not sure how to solve it.
Here is my thoughts:
- Write down the characteristic equations.
- Solve the characteristic equations.
- Use the solutions to rewrite the original PDE in canonical form.
For 1, here is my attempt $\frac{dx}{dt}=1,\quad\frac{dy}{dt}=-2,\quad\frac{du}{dt}=9$, but I am not entirely sure and don't know how to proceed further.
To solve the PDE $$ (x-2y+9)u_x+(y+9)u_y=9u \tag{1} $$ using the method of characteristics, we first write the Lagrange-Charpit equations in its parametrization invariant form: $$ \frac{dx}{x-2y+9}=\frac{dy}{y+9}=\frac{du}{9u}. \tag{2} $$ The solution of $$ \frac{dy}{y+9}=\frac{du}{9u} \tag{3} $$ yields $$ u=C_1(y+9)^9. \tag{4} $$ The solution of $$ \frac{dx}{x-2y+9}=\frac{dy}{y+9} \tag{5} $$ is slightly more complicated. Let's first rewrite it in terms of the variable $z=y+9$: $$ \frac{dx}{x-2z+27}=\frac{dz}{z}\implies \frac{dx}{dz}-\frac{x}{z}=\frac{27}{z}-2. \tag{6} $$ Dividing both sides of $(6)$ by $z$ we obtain $$ \frac{1}{z}\frac{dx}{dz}-\frac{x}{z^2}=\frac{d}{dz}\left(\frac{x}{z}\right)= \frac{27}{z^2}-\frac{2}{z} $$ $$ \implies \frac{x}{z}=-\frac{27}{z}-2\ln|z|+C_2 \implies \frac{x+27}{y+9}+2\ln|y+9|=C_2, \tag{7} $$ where we replaced $z$ with $y+9$ in the last equation. Combining $(4)$ and $(7)$ we finally obtain the general solution to $(1)$: $$ u(x,y)=(y+9)^9F\left(\frac{x+27}{y+9}+2\ln|y+9|\right), \tag{8} $$ where $F$ is an arbitrary differentiable function.