# Simple PDE: general solution?

1. May 20, 2014

### quantum_smile

1. The problem statement, all variables and given/known data
Find the general solution of
$$u_{xx} + u = 6y,$$
in terms of arbitrary functions.

2. Relevant equations

The PDE has the homogeneous solution, $$u(x,y)=Acos(x)+Bsin(x)$$.
$$u_{xx} + u = 6y$$ has the particular solution, $$u(x,y)=6y$$

3. The attempt at a solution

Taking a superposition of the homogeneous and particular solutions, we can write that
u(x,y)=Acos(x)+Bsin(x)+6y. <--my solution

My question is, how do I know whether or not this is *the* most general solution? How do I know that I haven't missed something?

2. May 20, 2014

### Zondrina

You know that $y_h$ and $y_p$ are both linearly independent solutions to the equation. Therefore the sum of those two linearly independent solutions is also a linearly independent solution. Hence the most general solution (although not necessarily unique) is $y = y_h + y_p$.

Consider a regular ODE for a moment of the form: $a(x)y'' + b(x)y' + c(x)y = f(x)$.

Differentiating $y$ twice and plugging it into the above equation you will find the answer to be $0 + f(x) = f(x)$.

3. May 20, 2014

### LCKurtz

You don't know that it is. And obviously it isn't because you could have$$u(x,y) = A(y)\cos x + B(y)\sin x + 6y$$where $A(y)$ and $B(y)$ are arbitrary functions of $y$. Is this the most general solution? It might be, but I don't know. Usually problems like this come with some boundary conditions and assumptions which guarantee a unique solution. When you have that situation, then if you come up with something that works, no matter how you found it, you know you are done because you have the only solution there is.