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That is, given an equation

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

I know that solutions are of the form $e^{rt},$ where r is a constant computed from $ar^2 + br + c = 0$. For some reason, I have written down in my notes adjacently that the "homogenous solution" is $c_1e^{r_1t}+c_2e^{r_2t}$. I also didn't note what precisely that was, but my assumption was that it means any equation of the form $c_1e^{r_1t}+c_2e^{r_2t}$ also satisfies the initial equation.

Is the homogenous solution really valid for $ay''+by'+cy = 0$ in all places? If so, why is the starting point that the solutions are of the form $e^{rt}$, and why is the homogenous solution valid anyway?

If not, what is meant by the "homogenous solution"?

SquarerootSquirrel
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Homogeneous solutions of differential equations are solutions to the problem you stated, i.e. linear, constant coefficient equations with a zero right hand side. One way to think of it is that you have a mathematical operator

$$ Ly = ay^{\prime\prime} + by^\prime + cy $$ This operator takes in functions, computes the appropriate derivatives and multiplies by the constants, then spits out another function. This operator will have zero functions, just like a quadratic equation has roots. Those zero functions, i.e. the functions $y(t)$ such that $Ly$ is the zero function are called homogeneous solutions. They solve

$$ Ly = 0 $$

The reason why we distinguish them is that you might want to solve an inhomogeneous problem, i.e. one that looks like

$$ Ly = f(t) $$ This equation will have a solution that looks like $y = y_h+y_p$, where $y_h$ is a homogeneous solution (solves $Ly=0$) and $y_p$ is a solution to the full equation $Ly = f$. If you've taken linear algebra, a homogeneous solution is the same idea as a vector in the null space of a matrix.

The reason we always use $e^{rt}$ to "find" homogeneous solutions is that exponential functions are "eigenfunctions" of the derivative. This is just a fancy way of saying that if $y(t) = e^{rt}$,

$$ \frac{d}{dt}y(t) = re^{rt} = ry(t) $$ so the derivative of an exponential is a multiple of the same exponential. This means that dropping $e^{rt}$ into $L$ will convert the differential equation into a quadratic equation for $r$!

Will a solution of the form $e^{rt}$ always work? Yes, if you allow $r$ to be a complex number. This is because all quadratic polynomials have 2 complex roots (counting multiplicity).

icurays1
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An $n$th order, linear homogenous ODE has $n$ linearly independent solutions (See this for why.). For a $2$nd order linear homogenous ODE, we then expect two solutions.

In your case, the coefficients of the ODE are contants. This arises to the notable family of solutions that you've noted: $c_1e^{r_1t}+c_2e^{r_2t}$.

You've likely seen this, but let's assume $y=e^{rt}$. Upon substitution into the ODE, we can see our assumption is satisfied and indeed this is one of many solutions:

$$ay''+by'+cy = 0$$ $$a(e^{rt})''+b(e^{rt})'+ce^{rt} = 0$$ $$ar^2e^{rt}+bre^{rt}+ce^{rt} = 0$$ $$ar^2+br+c = 0$$

Note that the division by $e^{rt}$ is permitted since $e^{rt}>0$. This last line, though, is just a quadratic equation with roots:

$$r = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

The two roots give $r_1$ and $r_2$, and thus let us construct a general solution. Since it is linear, we sum the two solutions (which happen to be linearly independent. Look up Wronksian for further info). The summation of solutions is a solution itself, hence

$$y = c_1e^{r_1t}+c_2e^{r_2t}$$

The above is your real-valued solution for DISTINCT roots $r_1 \ne r_2$.

But what if $r_1 = r_2$ in the case of repeated roots? Or $r_1, r_2 \in \mathbb{C}$ in the case of complex roots? There are workarounds for these cases, and the answer will be again represented as a sum of two linearly independent solutions.

zahbaz
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