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

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

So our equation reads

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

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.

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October 10, 2008 - Posted by John Armstrong | Analysis, Calculus, Differential Equations

3 Comments »

[...] and Cosine Now I want to consider the differential equation . As I mentioned at the end of last time, we can write this as and find two solutions — and — by taking the two complex [...]

Pingback by Sine and Cosine « The Unapologetic Mathematician | October 13, 2008 | Reply
[...] can be used to express analytic functions. Then I showed how power series can be used to solve certain differential equations, which led us to defining the functions sine and cosine. Then I showed that the sine function must [...]

Pingback by Pi: A Wrap-Up « The Unapologetic Mathematician | October 16, 2008 | Reply
Congratulation Sir. The power series has been commonly used to solve second order ODE. But, here you have explored to solve the first order ODE. I obtain so many information from you.

Comment by rohedi | November 29, 2008 | Reply

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