Determine the values for r in the differential equation

**VitaX** · August 30th 2011, 04:35 PM

Determine the value of r for which the given differential equation has solutions of the form $y=t^r$ for t > 0.

$t^2 y'' + 4ty' + 2y = 0$

$y'(t) = rt^{r-1}$

$y''(t) = r(r-1)t^{r-2}$

$t^2[r(r-1)t^{r-2}] + 4t[rt^{r-1}] + 2t^r = 0$

I'm a little confused as how to simplify this down. Can $t^2$ multiplied by $t^{r-2}$ produce $t^r$ ?

**TKHunny** · August 30th 2011, 05:16 PM

Rules of exponents. Why do you hesitate? Go forth and serve.

**VitaX** · August 30th 2011, 05:32 PM

I had forgotten is all, haven't had a math course in a while so I was rusty and for some reason the r constant was throwing me off.

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