Partial differential equation : $z_x+z_y+z=e^{x+2y},z(x,0)=0$

Question

How to solve this equation: $$z_x+z_y+z=e^{x+2y}$$ with boundary condition as $z(x,0)=0$

I tried $$\frac{dx}{1}=\frac{dy}{1}=\frac{dz}{e^{x+2y}-z}$$ I got one condition as $y-x=a$ how to obtain other constant condition.? can you please help ? How to fix that extra z or exponential term rid of dz, I felt we should use some mulitplier tricks ?

Answer 1

Solution of $z_x+z_y+z=0$ is $$z_h=e^{-x}f(y-x)$$

Particular solution of $z_x+z_y+z=e^{x+2y}$ is $$ z_p=\frac{e^{x+2y}}{4}$$ Then general solution is $$z=z_h+z_p=e^{-x}f(y-x)+\frac{e^{x+2y}}{4}$$

Final solution with initial condition $z(x,0)=0$ is $$z=\frac{e^{x+2y}}{4}-\frac{e^{x-2y}}{4}=\frac12e^x\sinh(2y)$$

Answer 2

Using $r=\frac ab=\frac cd\implies r=\frac{a+c}{b+d}$ gives as second relation $$ \frac{dx+2dy}{1+2}=\frac{dz}{e^{x+2y}-z} $$ so that with $r=x+2y$ we get $$ \frac{dz}{dr}=\frac{e^r-z}3 \\ \frac{d}{dr}(e^{r/3}z)=\frac{e^{4r/3}}3 \\ e^{r/3}z=c_2+\frac{e^{4r/3}}4 $$ so that going back to the general procedure with $c_2=\phi(c_1)$ it follows that $$ z=\phi(y-x)e^{-(x+2y)/3}+\frac14e^{x+2y} $$ Now insert the initial condition $$ 0=z(-x,0)=\phi(x)e^{x/3}+\frac14e^{-x}\implies \phi(x)=-\frac14 e^{-4x/3} $$ and combine to get the solution to this specific problem.