Help solving a partial differential equation by separation of variables

Question

Here is thtial x} - 5y\fal u}{\partial x} - 5y\frac{\partial u}{\partial y} + 4u=0$$

I have isolated tial x} - 5y\f $$\frac{x^2}F\tial x} - 5y\fl x} - 5y\fG}{dy}$$where we assume $u$ is separable i.e. $u(x,y)=F(x)G(y)$

I am stumped on where to go from here.

Edit: tial x} - 5y\fy}G\frac{dG}{dy}=λ$$

So now can I tial x} - 5y\flve these seperately? How do I go about doing this?

Answer 1

Now you can see that the first member depends only on the variable $s$ and the second member depends only on the variable $y$.

So it's clear that to the equality holds the first member has to be a constant, such as the second;

Indicating the first member as $A$ and the second as $B$, we have:

$$\frac{x^2}{f(x)} \frac{d^2f(x)}{dx^2} + \frac{5x}{f(x)} \frac{df(x)}{dx}+4=A$$

$$\frac{x^2}{f(x)} \frac{d^2f(x)}{dx^2} + \frac{5x}{f(x)} \frac{df(x)}{dx}+4=B$$

And you can propose a solution of the kind $f(x,y)=F(x)G(y)$, by making a separation of the variables. With this, you can solve the both equations independently (I guess).

Answer 2

First, note we assume our solution is separable i.e. $u(x,y)=F(x)G(y)$. Let's jump to where you left off.

Since the left-hand side of our equality depends only on $x$ and the right-hand side only on $y$ it follows both equations are equal to some constant, say $\lambda$:$$x^2\frac{d^2F}{dx^2}+5x\frac{dF}{dx}+(4-\lambda)F=0\\y^2\frac{d^2G}{dy^2}-5y\frac{dG}{dy}+\lambda G=0$$Both of these ordinary differential equations are Cauchy-Euler and so yield simple-to-find solutions; with ansatz $F(x)=x^n$ we have:$$n(n-1)x^n+5nx^n+(4-\lambda)x^n=0\\n(n-1)+5n+(4-\lambda)=0\\n^2+4n+4=\lambda\\(n+2)^2=\lambda\\n=-2\pm\sqrt{\lambda}$$giving the general solution $F(x)=c_1x^{-2+\sqrt\lambda}+c_2x^{-2-\sqrt\lambda}$ due the linearity of the problem.

A similar ansatz for our second equation, $G(y)=y^m$, yields:$$m(m-1)y^m-5my^m+\lambda y^m=0\\m^2-6m+\lambda=0\\m^2-6m+9=9-\lambda\\(m-3)^2=9-\lambda\\m=3\pm\sqrt{9-\lambda}$$giving similarly $G(y)=c_3y^{3+\sqrt{9-\lambda}}+c_4y^{3-\sqrt{9-\lambda}}$.

Ultimately this gives the solution:$$u(x,y)=(c_1x^{-2+\sqrt\lambda}+c_2x^{-2-\sqrt\lambda})(c_3y^{3+\sqrt{9-\lambda}}+c_4y^{3-\sqrt{9-\lambda}})=\dots$$