To solve a PDE by separation of variables $\displaystyle 4\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=3u$

Question

To Solve: $\displaystyle 4\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=3u$

My attempt:

Let $\displaystyle u=X(x)Y(y)$. So, $\displaystyle 4X'Y +XY'=3XY$

Separating the variables, $\displaystyle \frac{4X'-3X}{X}=\frac{-Y'}{Y}=k$(say)

$\displaystyle X'-\frac{(3+k)}{4}X=0$

$\displaystyle Y'+kY=0$

Integrating, $\displaystyle X=c_1e^{(k+3)x/4}$

and $\displaystyle Y=c_2e^{-ky}$

So, $\displaystyle u=c_1c_2e^{(k+3)x/4-ky}$

Now I check this against the initial value given as $\displaystyle u=3e^{-y}-e^{5y}$ when $x=0$

The final answer also (obviously) has two components $u=3e^{x-y}-e^{2x-5y}$

How come there are two components here? What did I miss?

Answer 1

I don't have time to type it out in Mathjax, I have taken an image of the solution and please pardon me if it is not clear. Thanks

Answer 2

As the PDE is linear and homogeneous, a linear combination of two solutions is another solution. So, even though your assumed solution is of the form $X(x)Y(y)$, you can get a solution of the form $X_1(x)Y_1(y) + X_2(x)Y_2(y)$.

The two terms in the solution are each separate solutions.

Why is the sum of two solutions again a solution? Suppose $u_1$ and $u_2$ two solutions. Then $4\dfrac{\partial (u_1 + u_2)}{\partial x} + \dfrac{\partial (u_1 + u_2)}{\partial y} = 4\dfrac{\partial u_1}{\partial x} + \dfrac{\partial u_1}{\partial y} + 4\dfrac{\partial u_2}{\partial x} + \dfrac{\partial u_2}{\partial y} = 3u_1 + 3u_2 = 3(u_1 + u_2)$, so $u_1 + u_2$ is also a solution.