Consider the PDE $y^{2}*u_x + x^{2}u_y=0$ Derive the product solutions

Question

Consider the PDE $$y^{2}u_x + x^{2}u_y=0$$ Derive the product solutions for u = u(x,y) and for unknown separation constant λ and let C denote some arbitary constant.

This is what I have so far:

---Define u = XY ---

---Sub into our original equation ---

$$ y^2 X'Y + x^2XY' = 0$$

--- Divide both sides by XY ---

$${y^2X' \over X} + {x^2Y' \over Y} = 0$$

---Rearranging---

$${y^2X' \over X} = - {x^2Y' \over Y} = \lambda$$

$${X' \over x^2X} = - {Y' \over y^2Y} = \lambda$$

---Split into two separate equations since we know $\lambda$ is constant ---

Equation 1:

$$ {X' \over (x^2X)} = \lambda$$

Equation 2:

$$ - {Y' \over (y^2Y)} = \lambda$$

This is where I am stuck. I don't know how to solve these two separate equations. Any help would be appreciated.

Answer 1

$X^\prime = \frac{dX}{dx}$ and $Y^\prime = \frac{dY}{dy}$. Then separate $X$ and $x$ in the first, and $Y$ and $y$ in the second.

Answer 2

$$ {X' \over (x^2X)} = \lambda \quad\implies\quad \frac{dX}{X}=\lambda x^2dx \quad\implies X=a\:e^{\frac{\lambda}{3}x^3}$$

$$ {Y' \over (y^2Y)} = -\lambda \quad\implies\quad \frac{dY}{Y}=\lambda y^2dy \quad\implies Y=b\;e^{\frac{\lambda}{3}y^3}$$

$$u(x,y,\lambda)=XY=c\:e^{\frac{\lambda}{3}x^3}e^{\frac{\lambda}{3}y^3} =c\: e^{\frac{\lambda}{3}(x^3-y^3)}$$

There are as many solutions $u(x,y,\lambda)$ as they are different $\lambda$ as well as any linear combination between them. So the general solution can be expressed as : $$u(x,y)=\int c(\lambda)e^{\frac{\lambda}{3}(x^3-y^3)}d\lambda$$

where $c(\lambda)$ is any function of $\lambda$.

In the above integral, the variable for integration is $\lambda$ only. So, $k=(x^3-y^3)$ can be considered as a parameter. Since the factor $c(\lambda)$ is any function of $\lambda$, the result of integration is any value, but depending of $k$. In other words, the result is any function of $k$. $$u(x,y)=F(x^3-y^3)$$ where $F$ is an arbitrary function.

This is consistent with the solution from the method of characteristics. In short :

The characteristic ODEs are : $\quad \frac{dx}{y^2}=\frac{dy}{x^2}=\frac{du}{0}$

A first family of characteristic curves comes from $\quad \frac{dx}{y^2}=\frac{dy}{x^2} \quad\implies\quad x^3-y^3=c_1$

A second family of characteristic curves comes from $\quad \frac{du}{0}=$finite $\quad\implies\quad u=c_2$

The general solution of the PDE expressed on the form of implicit equation is : $$\Phi\left((x^3-y^3)\:,\: u \right)=0$$ where $\Phi$ is any function of two variables.

Or equivalently on explicit form : $$u(x,y)=F(x^3-y^3)$$ where $F$ is any function.

This is the same result obtained above with the method of separation of variables.