Help with second order initial value problem

**steph3824** · March 8th 2010, 05:32 PM

$y''+8y'-9y=0, y(1)=1, y'(1)=0$

I have found that the two roots for this are $r_1=-9, r_2=1$

So, $y(t)=c_1e^{-9t}+c_2e^t$
$y'(t)=-9c_1e^{-9t}+c_2e^t$

After this, I get pretty lost trying to get the correct values for c1 and c2. Can someone help out with this part please? I know that the ans. to this problem is $y=1/10e^{9-9t}+9/10e^{-1+t}$

**pickslides** · March 8th 2010, 05:41 PM

Originally Posted by steph3824

$y''+8y'-9y=0, y(1)=1, y'(1)=0$

So, $y(t)=c_1e^{-9t}+c_2e^t$
$y'(t)=-9c_1e^{-9t}+c_2e^t$

Use $y(1)=1, y'(1)=0$

So,

$y(1)=c_1e^{-9\times 1}+c_2e^1 = 1$

and

$y'(1)=-9c_1e^{-9\times 1}+c_2e^1= 0$

Now simplify and solve.

**steph3824** · March 9th 2010, 05:57 PM

Well that's what I'm needing help with is the simplifying. I think that $c_2=1-c_1e^{-9}/e$ but I'm not even positive this is correct, does it become $c_2=1-c_1e^{-10}$ ?

So if the first thing is correct you get $-9c_1e^{-9}+(1-c_1e^{-9}/e)e=0$ and then $-9c_1e^{-9}+1-c_1e^{-9}=0$

I don't really know if that is correct and if it is how do you get c_1? I've forgotten the exponent rules and what not

**pickslides** · March 9th 2010, 06:02 PM

I get $c_2 = \frac{-9c_1}{e^{10}}$

**steph3824** · March 9th 2010, 06:03 PM

Nevermind, after working with the numbers a bit I have figured it out

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