## 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 $y^{\prime\prime\prime}-y^{\prime} = 0$.