# The Unapologetic Mathematician

## First-Degree Homogeneous Linear Equations with Constant Coefficients

Now that we solved one differential equation, let’s try a wider class: “first-degree homogeneous linear equations with constant coefficients”. Let’s break this down.

• First-degree: only involves the undetermined function and its first derivative
• Homogeneous: harder to nail down, but for our purposes it means that every term involves the undetermined function or its first derivative
• Linear: no products of the undetermined function or its first derivative with each other
• Constant Coefficients: only multiplying the undetermined function and its derivative by constants

Putting it all together, this means that our equation must look like this:

$\displaystyle a\frac{df}{dx}+bf=0$

We can divide through by $a$ to assume without loss of generality that $a=1$ (if $a=0$ then the equation isn’t very interesting at all).

Now let’s again assume that $f$ is analytic at ${0}$, so we can write

$\displaystyle f(x)=\sum\limits_{k=0}^\infty a_kx^k$

and

$\displaystyle f'(x)=\sum\limits_{k=0}^\infty (k+1)a_{k+1}x^k$

$\displaystyle0=\sum\limits_{k=0}^\infty\left((k+1)a_{k+1}+ba_k\right)x^k$

That is, $a_{k+1}=\frac{-b}{k+1}a_k$ for all $k$, and $a_0=f(0)$ is arbitrary. Just like last time we see that multiplying by $\frac{1}{k+1}$ at each step gives $a_k$ a factor of $\frac{1}{k!}$. But now we also multiply by $(-b)$ at each step, so we find

$\displaystyle f(x)=\sum\limits_{k=0}^\infty\frac{(-b)^kx^k}{k!}=\sum\limits_{k=0}^\infty\frac{(-bx)^k}{k!}=\exp(-bx)$

And indeed, we can rewrite our equation as $f'(x)=-bf(x)$. The chain rule clearly shows us that $\exp(-bx)$ satisfies this equation.

In fact, we can immediately see that the function $\exp(kx)$ will satisfy many other equations, like $f''(x)=k^2f(x)$, $f'''(x)=k^3f(x)$, and so on.

October 10, 2008 -