## Tuesday, September 12, 2006

### Multiply By Something. Topic: Calculus.

Problem: Solve the differential equation $y^{\prime\prime}+3y^{\prime}-4y = 0$.

Solution: Let's conveniently rewrite this as

$y^{\prime\prime}+4y^{\prime} = y^{\prime}+4y$

and multiply by something, perhaps $e^{4x}$. And then integerate, which is

$\displaystyle \int (y^{\prime\prime}+4y^{\prime})e^{4x} = \int (y^{\prime}+4y)e^{4x}$.

Conveniently enough, we see that this is

$y^{\prime}e^{4x} = ye^{4x}+C_1$

(you can verify this by taking the derivative). Well we might as well be awesome and try that trick again. Multiply the whole thing by $e^{-5x}$ now and integrate

$\displaystyle -\int C_1 e^{-5x} = \int (y-y^{\prime})e^{-x}$.

The LHS is easy and integrating the RHS makes the equation

$\frac{1}{5}C_1e^{-5x} = -ye^{-x}+C_2$

(again, check by differentiating). Now solving for $y$, we have

$y = -\frac{1}{5}C_1 e^{-4x}+C_2 e^x = Ae^{-4x}+Be^x$

as the general solution. QED.

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Comment: A cool method for solving differential equations; it actually works for a lot of them, but obviously doesn't work for a lot, too. In any case, it took me some time to develop this method during class.

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Practice Problem: Solve the differential equation $y^{\prime\prime\prime}-y^{\prime} = 0$.

#### 2 comments:

1. linear autonomous differential equations all translate into some recurrence involving the taylor polynomial basically

if you can solve that recurrence, which is often just an application of the characteristic edquation concept, you should be able to find the function; for these examples, you'll end up with a second-order linear homogeneous recurrence whose solutions are a sum of two geometric series, ergo, the original function is a sum of two exponentials

2. Yay, calculus. You need to put more of this stuff on. And I actually get it.