# Not getting the right answer in Reduction of Order Problem

• May 28th 2011, 01:27 PM
Carbon
Not getting the right answer in Reduction of Order Problem
$\displaystyle x''=\frac{t+1}{t}x'-\frac{1}{t}x$

with $\displaystyle x_1(t)=e^t$

The goal is to find the other solution.

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

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

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

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

but the answers supposed to be $\displaystyle t-1$ where did I go wrong? thanks
• May 28th 2011, 02:47 PM
Jester
Why do you think you're wrong? If you substitute your answer into the original ODE you get

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

which I believe identically satisfies your ODE!
• May 28th 2011, 03:16 PM
Carbon
Because of this...

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

So there are more than two solutions?
• May 28th 2011, 04:15 PM
Jester
If you look, there's a slight typo (on the negative sign).
• May 28th 2011, 04:28 PM
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...
• May 28th 2011, 04:42 PM
TheEmptySet
Quote:

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

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

It should be

$\displaystyle t+1$