# Solving second-order linear homogenous differential equations

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?

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

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.)

• 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? – Mihael Apr 6 at 19:22
• @Mihael: because $(e^x)'=e^x$. – Yves Daoust Apr 7 at 8:07