# Thread: Not getting the right answer in Reduction of Order Problem

1. ## 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

2. 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!

3. Because of this...

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

So there are more than two solutions?

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

5. 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...

6. 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

$\displaystyle t+1$