The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Analysis, Calculus, Differential Equations

3 Comments »

  1. [...] 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

  2. [...] 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

  3. 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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