# No-homo second order DE

1. Apr 29, 2013

### 1MileCrash

1. The problem statement, all variables and given/known data

y''-2y'+2y = e^-t, y(0)=0, y'(0)=4

2. Relevant equations

3. The attempt at a solution

Again.. I find it to be really easy, then get it wrong. My answer is not even close. Applying the properties of the laplace transform in the usual way,

s^2L{y} - 4 - 2L{y} + 2L{y} = L{e^-t}

=>

$L{y} = (\frac{1}{s+1})(\frac{1}{s^2 - 2s + 2}) + (\frac{4}{s^2-2s+2})$

Now, I split the first term into partial fractions and factored the denomenator of the second. I'll spare you all of that tedious algebra:

$-\frac{1}{4}(\frac{1}{s+1}) + \frac{1}{4}(\frac{1}{s-1}) - \frac{1}{2}(\frac{1}{(s-1)^{2}}) + 4(\frac{1}{(s-1)^{2}})$

The inverse laplace of that, I find to be

-(1/4)e^-t + (1/4)e^t - (1/2)te^t + 4te^t

The answer given and the one given by wolfram involves trig functions, what am I doing wrong??

Thanks!

2. Apr 29, 2013

### Staff: Mentor

For starters, s2 - 2s + 2 ≠ (s - 1)2.

3. Apr 29, 2013

### 1MileCrash

I think I need a nap.