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I'm trying to solve the differential equation above by separating the variables. Thus, I let $u(x,y) = X(x)Y(y)$. As such, $u_x = X'(x)Y(y)$ and $u_y = Y'(y)X(x)$.

The differential equation above can then be rewritten as: $X'(x)Y(y) + 2xX(x)Y'(y) + (2x - 6)X(x)Y(y) = 0$

However, I find myself stuck after this step as I'm unable to separate the variables and shift all the functions/variables with x to 1 side of the equation and all the functions/variables with y to the other side of the equation.

Some help will be greatly appreciated. Thanks a lot :)

Jenson Mutton
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1 Answers1

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$$u_x + 2xu_y = (6-2x)u$$ Why not the method of characteristics ?

https://en.wikipedia.org/wiki/Method_of_characteristics

The Charpit-Lagrange system of characteristic ODEs is : $$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{(6-2x)u}$$ A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{2x}$ : $$y-x^2=c_1$$ A second characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{2x}=\frac{6dx-dy}{6-2x}=\frac{du}{(6-2x)u}\quad;\quad d(6x-y)=\frac{du}{u}$ $$u\:e^{y-6x}=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is : $$u\:e^{y-6x}=F(y-x^2)$$ $F$ is an arbitrary function. $$\boxed{u(x,y)=e^{6x-y}F(y-x^2)}$$ If some boundary condition was specified in the wording of the problem one could determine the function $F$ in order to find the particular solution satisfying both the PDE and the boundary condition.

JJacquelin
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