Inital Value Problem with Laplace Transforms

**diddledabble** · July 25th 2009, 06:21 AM

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

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

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

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

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

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

**galactus** · July 25th 2009, 06:34 AM

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

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

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

**Random Variable** · July 25th 2009, 08:56 AM

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

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

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

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

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

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

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

**diddledabble** · July 25th 2009, 09:12 AM

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

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

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

**Random Variable** · July 25th 2009, 09:47 AM

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.

Thread: Inital Value Problem with Laplace Transforms

LinkBack

Thread Tools

Display

Inital Value Problem with Laplace Transforms

You lost me.

Similar Math Help Forum Discussions

LaPlace transforms!

Laplace Transforms to solve Initial-Value problem

Inital value problem

Laplace transforms

Laplace Transforms

Search Tags