If $y_1$ and $y_2$ are both solutions to the non-homogenous equation

$$ay'' + by' + cy = f(x)$$

is $y_1-y_2$ to solution to the homogenous equation

$$ay'' + by' + cy = 0?$$

So far this is what I have

$$ay_1'' + by_1' + cy_1' = f(x) = ay_2'' + by_2' + cy_2$$

$$ay_1'' + by_1' + cy_1 - (ay_2'' + by_2' + cy_2) = 0$$

$$a(y_1-y_2)'' + b(y_1-y_2)' + c(y_1-y_2) = 0 $$

And then I'm not sure about this step

$$(y_1-y_2)'' + (y_1-y_2)' + (y_1-y_2) = 0$$

Do I go about integrating (or deriving) sections of both sides until I end up with a $y_1 = y_2$ statement, therefore solving the homogenous equation? It feels as though I'm breaking some mathematical rule, even though by the FTOC I should be safe to derive or integrate each element of the equation.

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    $\begingroup$ Your work is fine. If $y_1$ and $y_2$ are solutions, then so is $\alpha y_1 + \beta y_2$ for any real numbers $\alpha,$ and $\beta.$ $\endgroup$ – Chickenmancer Apr 28 '17 at 0:34
  • $\begingroup$ Your third equation shows that $y_1 - y_2$ is a solution to the homogeneous equation $ay'' + by' + cy = 0$. If you stop there, you are done. The fourth equation is wrong and I don't understand what you are talking about in your last paragraph. $\endgroup$ – NickD Apr 28 '17 at 2:01



The interesting result in solving differential equations like these ones is that the general solution can be build in this way :

$\mathtt{general\ solution\ of\ (E)} = \mathtt{general\ solution\ of\ (H)} + \mathtt{particular\ solution\ of\ (E)}$

The homogeneous equation $(H)$ is easier to solve, and we need only $1$ particular solution of the full equation to get overall general solutions.

This is exactly what your question is about to prove, when you take $2$ general solutions of $(E)$ then their difference is solution of $(H)$

And you arrived at that result when you wrote : $a(y_1-y_2)''+b(y_1-y_2)'+c(y_1-y_2)=0$.

So you were in fact arrived at the result (i.e. $(y_1-y_2)$ is solution of $(H)$), but went astray in the mistaken belief that you had to prove $y_1=y_2$.


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