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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?

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  • $\begingroup$ $y = ce^{-\frac{b}{a}}$ 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
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The DE is

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

If you assume that the solution is of the form

$$y=e^{kx}$$

and you plug this solution in the equation, you obtain

$$ak^2e^{kx}+bke^{kx}+ce^{kt}=0.$$

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

$$ak^2+bk+c=0.$$

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

$$C_0e^{k_0x}+C_1e^{k_1x}.$$


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

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  • $\begingroup$ 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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