# Why is every solution to a homogenous second-order linear differential equation in the form $C_0e^{\alpha x} + C_1e^{\beta x}$

In textbooks, it's often casually mentioned, without explanation, that any two solutions added together is the general solution, the form of every other solution.

I don't understand why this is or where that idea comes from. Can anyone explain why this is so?

EDIT: This question was badly put. I'm going to post another question that makes more sense. Thanks for the feedback.

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Sometimes the general solution has the form $(C_0+C_1x)e^{\alpha x}$. –  Hagen von Eitzen Nov 25 '12 at 23:37

It's not true that every solution has the form given in the title of the question.

It is true that any linear combination of solutions is a solution or, to put the same thing another way, that the set of all solutions forms a vector space. Do you need help proving that the set of all solutions forms a vector space?

The only question, then, is why the vector space has dimension 2. Well, one proves a theorem stating that there is a unique solution given the initial conditions $y(0)$ and $y'(0)$, and that settles it.

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The premise in your title is incorrect, even for constant-coefficient equations (and certainly not for non-constant-coefficient ones).
For example, $y'' + 2 y' + y = 0$ has the solutions $y = C_0 e^{-x} + C_1 x e^{-x}$.

What is true is that by the Existence and Uniqueness Theorem and linearity, the space of solutions of a homogeneous second-order linear differential equation is a two-dimensional vector space. So if you have two linearly independent solutions, all solutions are linear combinations of these two.

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While excellent answers have already been given, I think you might like some more concrete steps. Say I have two solutions, $y_1$ and $y_2$. If the equation is represented by some linear differential operator $D$ (it could be $D=\frac{d^2}{dx^2}+x$, $D=x^2\frac{d^2}{dx^2}-\frac{d}{dx}$ or any number of things), then

$$D(y_1)=0$$ $$D(y_2)=0$$

these are zero precisely because $y_1$ and $y_2$ are both solutions to the homogeneous ODE by assumption. So clearly

$$D(y_1)+D(y_2)=0$$

and if the operator is linear (which it is here)

$$D(y_1+y_2)=0$$

So if you have two solutions, by linearity their sum is a solution as well. At its core this is just a fancy application of $0+0=0$. A very similar statement can be made about scalar multiples. Why two independent solutions are needed to find the rest (i.e, they form a basis) has been answered excellently by others.

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