What Is Wrong with My Solution for This Second Order Differential Equation?

[1MileCrash]

Homework Statement

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

Homework Equations

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!

 

[Mark44]

Homework Statement

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

Homework Equations

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!

For starters, s² - 2s + 2 ≠ (s - 1)².

 

[1MileCrash]

For starters, s² - 2s + 2 ≠ (s - 1)².

I think I need a nap.