I am stuck in understanding part of the explanation of solving second-order linear homogenous differential equations in my textbook.

We start with the first-order equation, where $a, b \in \mathbb{R}$

$ay' + by = 0$

Using the method involving the reverse of the Product rule, the solution is

$y = ce^{-\frac{b}{a}x}$, where $c$ is an arbitrary constant.

Since both $a$ and $b$ are constants to be deterimed, this is re-written using $k=-\frac{b}{a}$

$y = ce^{kx}$

Then, the explanation in the book is making a logical connection which I fail to see. The second-order equation that we are aiming to solve is $ay'' + by' + cy = 0$.

It says, since $y = ce^{kx}$ is the solution of $ay' + by = 0$, and because $k$ is the solution of $ak+b=0$, it follows that $y = ce^{kx}$ might be a solution of $ay'' + by' + cy = 0$ (although a non-general solution, as it has only one arbitrary constant).

I must be blind, but I cannot see the connection and how does $ak+b=0$ come into it?

  • $\begingroup$ $y = ce^{-\frac{b}{a}}$ is false. The correct equation is : $$y = ce^{-\frac{b}{a}x}$$ $\endgroup$ – JJacquelin Apr 6 at 15:49
  • $\begingroup$ @JJacquelin thanks, corrected it now. $\endgroup$ – Mihael Apr 6 at 15:53

The DE is


If you assume that the solution is of the form


and you plug this solution in the equation, you obtain


For this relation to hold for any $x$, you need


By the fundamental theorem of algebra, we know that this equation has two solutions, which are in general distinct. Let $k_0$ and $k_1$. Now by linearity of the equation,

$$e^{k_0x}$$ and $$e^{k_1x}$$ are two independent solutions, so that the general solution is of the form


In fact, this method works with linear equations of constant coefficients of any order. (I bypassed the case of multiple roots, for simplicity.)

  • $\begingroup$ Thanks, but my question is particularly about how do we assume that the solution must be in the form of $y=e^{kx}$ in the first place? I fail to follow the logical chain which attempts to explain this in my textbook. Can you help? $\endgroup$ – Mihael Apr 6 at 19:22
  • $\begingroup$ @Mihael: because $(e^x)'=e^x$. $\endgroup$ – Yves Daoust Apr 7 at 8:07

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