# IVP with Laplace Transform

• August 29th 2009, 09:28 PM
MathTragic
IVP with Laplace Transform
Solve the initial value problem using Laplace Transforms.

$\ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0$
• August 29th 2009, 11:32 PM
Chris L T521
Quote:

Originally Posted by MathTragic
Solve the initial value problem using Laplace Transforms.

$\ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0$

Apply the Laplace transform to both sides of the equation to get

$s^2Y(s)-s y(0)-\dot{y}(0)-6\left(sY(s)-y(0)\right)+9Y(s)=\frac{1}{(s-3)^2}$

Apply the initial conditions and simplify to get

$s^2Y(s)-6sY(s)+9Y(s)-s+6=\frac{1}{(s-3)^2}\implies (s-3)^2Y(s)=\frac{1}{(s-3)^2}+s-6$ $\implies Y(s)=\frac{1}{(s-3)^4}+\frac{s}{(s-3)^2}-\frac{6}{(s-3)^2}$

Take the Inverse Laplace Transform of both sides. I leave it for you to continue from here.

Spoiler:
For the first and third term on the RHS, use a translation theorem to make life easier. The second term on the RHS can be rewritten as $\frac{1}{s-3}+\frac{3}{(s-3)^2}$. Apply translation theorems when needed.
• August 30th 2009, 09:15 PM
wik_chick88
Quote:

Originally Posted by Chris L T521
Apply the Laplace transform to both sides of the equation to get

$s^2Y(s)-s y(0)-\dot{y}(0)-6\left(sY(s)-y(0)\right)+9Y(s)=\frac{1}{(s-3)^2}$

Apply the initial conditions and simplify to get

$s^2Y(s)-6sY(s)+9Y(s)-s+6=\frac{1}{(s-3)^2}\implies (s-3)^2Y(s)=\frac{1}{(s-3)^2}+s-6$ $\implies Y(s)=\frac{1}{(s-3)^4}+\frac{s}{(s-3)^2}-\frac{6}{(s-2)^2}$

Take the Inverse Laplace Transform of both sides. I leave it for you to continue from here.

Spoiler:
For the first and third term on the RHS, use a translation theorem to make life easier. The second term on the RHS can be rewritten as $\frac{1}{s-3}+\frac{3}{(s-3)^2}$. Apply translation theorems when needed.

is the $-\frac{6}{(s-2)^2}$ ie. the third term of the $Y(s)$ equation supposed to be $-\frac{6}{(s-3)^2}$ ?
• August 30th 2009, 10:29 PM
Chris L T521
Quote:

Originally Posted by wik_chick88
is the $-\frac{6}{(s-2)^2}$ ie. the third term of the $Y(s)$ equation supposed to be $-\frac{6}{(s-3)^2}$ ?

Ah..yes it is. Thank you for pointing out that typo. It's fixed now.
• September 2nd 2009, 09:26 PM
wik_chick88
i get the answer to be:
$y(t) = e^{3t} (2t + 1)$
is this correct?
• September 2nd 2009, 10:56 PM
CaptainBlack
Quote:

Originally Posted by wik_chick88
i get the answer to be:
$y(t) = e^{3t} (2t + 1)$
is this correct?

Does it satisfy the initial conditions?

CB