Not getting the right answer in Reduction of Order Problem

**Carbon** · May 28th 2011, 01:27 PM

$x''=\frac{t+1}{t}x'-\frac{1}{t}x$

with $x_1(t)=e^t$

The goal is to find the other solution.

my attempt
$x_2=ux_1$
$x_2'=e^{t}u'+e^{t}u$
$x_2''=e^tu''+2e^tu'+e^tu$

Substituting into the original equation I get...
$u''=-\frac{t-1}{t}u'$

substituting with u'' with v'...
$v'=-\frac{t-1}{t}v$
I integrate to
$ln v=ln t-t+c1$
$v=t\cdot e^{-t}=u'$
integrating again...
$u=(-t-1)e^{-t}$

$x_2=(-t-1)e^{-t}\cdot e^t$
$x_2=-t-1$

but the answers supposed to be $t-1$ where did I go wrong? thanks

**Danny** · May 28th 2011, 02:47 PM

Why do you think you're wrong? If you substitute your answer into the original ODE you get

$0 = \dfrac{t+1}{t} (-1) - \dfrac{1}{t} \left(-t-1\right)$

which I believe identically satisfies your ODE!

**Carbon** · May 28th 2011, 03:16 PM

Because of this...

http://www.mathhelpforum.com/math-he...wn-181831.html

So there are more than two solutions?

**Danny** · May 28th 2011, 04:15 PM

If you look, there's a slight typo (on the negative sign).

**Carbon** · May 28th 2011, 04:28 PM

Who has the typo? On the maple I only see that he moved the right part of the equation to the left side which seems like it's allowed...

**TheEmptySet** · May 28th 2011, 04:42 PM

Originally Posted by Carbon

Who has the typo? On the maple I only see that he moved the right part of the equation to the left side which seems like it's allowed...

In my finial line in the other thread I wrote

$e^{t}e^{-t}(t+1)=t-1$

It should be

$t+1$

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