# Thread: Inital Value Problem with Laplace Transforms

1. ## Inital Value Problem with Laplace Transforms

$y" + 6y' +5y = 12e^t$

Solve with the initial conditons y(0)=-1 y'(0)=6

$L{y"}s +6L{y'}s + 5Y(s)= 12e^t$

I am stuck on this because of the $12e^t$

The other problems I think I have done right.
Example:
y" -y' - 2y =0 y(0)=-2 y'(0)=5

$L{y'}(s)=sY(s)-y(0) = sY(s) +2$
$L{y"}(s) = s^{2}Y(s) -sy10-y'(0) = s^{2}Y(s) + s -5$

$(s^{2}Y(s) +s-5) - (sY(s) +2) -2Y(s) = 0$
$Y(s)(s^{2}-s-2)+s=0$

$Y(s) = \frac{-s}{s^{2}-s-2}$

$Y(s) = \frac{-s}{(s-2)(s+1)}$

2. Use a LaPlace on the $12e^{t}$

Which is $\frac{12}{p-1}$

Then, solve your equation for Y in terms of p.

3. $y'' +6y'+5y=12e^{t}$

$s^{2}Y(s)-sy(0)-y'(0) + 6sY(s) - 6y(0) + 5Y(s) = \frac {12}{s-1}$

$s^{2}Y(s) +s - 6 + 6sY(s) + 6 + 5Y(s) = \frac{12}{s-1}$

$Y(s)\big(s^{2}+6s+5\big) = \frac {12}{s-1} -s$

$Y(s)\Big((s+5)(s+1)\Big) = \frac{12+s- s^{2}}{s-1}$

$Y(s) = \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{1}{s-1} -\frac{3}{4} \frac{1}{s+5} -\frac{5}{4} \frac{1}{s+1}$

$y(t) = e^{t} - \frac{3}{4}e^{-5t} - \frac{5}{4}e^{-t}$

4. ## You lost me.

$Y(s) = \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{1}{s-1} -\frac{3}{4} \frac{1}{s+5} -\frac{5}{4} \frac{1}{s+1}$

You lost me when you seperated out the fractions. How did you do that?

$y(t) = e^{t} - \frac{3}{4}e^{-5t} - \frac{5}{4}e^{-t}$

5. Originally Posted by diddledabble
$Y(s) = \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{1}{s-1} -\frac{3}{4} \frac{1}{s+5} -\frac{5}{4} \frac{1}{s+1}$

You lost me when you seperated out the fractions. How did you do that?

$y(t) = e^{t} - \frac{3}{4}e^{-5t} - \frac{5}{4}e^{-t}$
$\frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{A}{s-1} + \frac{B}{s+5} + \frac{C}{s+1}$

$12+ s- s^2 = A(s+5)(s+1) + B(s-1)(s+1) + C (s-1)(s+5)$

You could expand and equate like terms to find A, B, and C. Or you could let s=1, -1, and -5.