Help with solving first order PDE with characteristic method

Question

Take the following PDE:

$$u_x-u_y+(x+y)u=0$$

Express it in standard form,

$$u_x-u_y=-(x+y)u$$

From here, form the characteristic system of equations,

$$\frac{dx}{dt}=1$$ $$\frac{dy}{dt}=-1$$ $$\frac{du}{dt}=-(x+y)$$

Where the first two equations are separable, and yield the following solutions: $$x=t+f_1(s)$$ $$y=-t-f_2(s)$$ Here I tried to get an expression for the parameter $s$ equating both expressions for $t$: $$x-f_1(s)=-y-f_2(s)$$ $$x+y=f_1(s)-f_2(s)$$ $$x+y=f_3(s)$$ $$s=f_3^{-1}(x+y)$$ With this at hand, I proceeded to solve the last equation as follows: $$\frac{du}{dt}=-(x+y)$$ $$\frac{du}{dt}=-(t+f_1(s)-(t+f_2(s)))$$ $$\frac{du}{dt}=-(f_1(s)-f_2(s))$$ $$\frac{du}{dt}=-f_3(s)$$ $$\frac{du}{dt}=f_4(s)$$ which yields the following solution: $$u=f_4(s)(t+f_5(s))$$ $$u=f_4(s)t+f_4(s)f_5(s))$$ $$u=f_4(s)t+f_6(s)$$ substitute now one of the relations for $t$: $$u=f_4(s)(x-f_1(s))+f_6(s)$$ $$u=f_4(s)x-f_4(s)f_1(s)+f_6(s)$$ $$u=f_4(s)x-f_7(s)+f_6(s)$$ $$u=f_4(s)x+f_8(s)$$ now substitute the value for $s$ to get the final solution, $$u=f_4(f_3^{-1}(x+y))x+f_8(f_3^{-1}(x+y))$$ $$u=M(x+y)x+N(x+y)$$

This is all well and good but the verification fails, I have traced back my steps but I can't find where I am wrong, and to add insult to injury the teacher told us that the solution should have a term $e^{x+y}$ for which I just can't find a way to get it. What am I missing? Any insight will be greatly appreciated.

Answer 1

Your solution needs two corrections:

1. Your fifth equation should be $$ \frac{du}{dt}=-(x+y)\,u. \tag{1} $$ Defining $v=\ln u$, it can be rewritten as $$ \frac{dv}{dt}=-(x+y), \tag{2} $$ so almost all of your subsequent equations remain valid if you replace $u$ with $v$.
2. In your next to last equation, you must recall that $f_4$ was previously defined as $-f_3$, so the last equation should be (replacing $u$ with $v$) $$ v=-x(x+y)+N(x+y). \tag{3} $$ Finally, since $v=\ln u$, we may rewrite $(3)$ as $$ u(x,y)=f(x+y)\,e^{-x(x+y)}, \tag{4} $$ where $f$ is an arbitrary differentiable function.

Answer 2

$$u_x-u_y=-(x+y)u$$

$$\frac{dx}{dt}=1\quad\text{OK.}$$ $$\frac{dy}{dt}=-1\quad\text{OK.}$$ $$\frac{du}{dt}=-(x+y)\quad\text{isn't correct. It should be}\quad \frac{du}{dt}=-(x+y)u$$ Equivalent writting : $$\frac{dx}{1}=\frac{dy}{-1}=\frac{du}{-(x+y)u}=dt$$ A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{-1}$ : $$x+y=c_1$$ A second characteristic equation comes from solving

$\frac{dx}{1}=\frac{dy}{-1}=\frac{x\,dx+(-y)\,dy}{x-(-y)}=\frac{du}{-(x+y)u}\quad\implies\quad \frac12 d(x^2-y^2)=\frac{du}{-u}$ $$u\:e^{\frac12(x^2-y^2)}=c_2$$ The general solution of the PDE on implict form $c_2=F(c_1)$ is : $$u\:e^{\frac12(x^2-y^2)}=F(x+y)$$ $F$ is an arbitrary function (to be determined according to some boundary condition). $$\boxed{u(x,y)=e^{\frac12(y^2-x^2)}F(x+y)}$$

NOTE :

The above result is consistent with the result from Gonçalo : $$e^{\frac12(y^2-x^2)}F(x+y)=e^{-x(x+y)}f(x+y)\quad\text{with}\quad f(x+y)=e^{\frac12(x+y)^2}F(x+y)$$