Question on the applications of Laplace transforms

**loolo** · October 5th 2012, 10:33 AM

hi

I need your help to resolve this question

y''-y'-2y=18e^-t sin3t

y(c)=0 , y'(c)=3

?

**TheEmptySet** · October 5th 2012, 10:47 AM

Originally Posted by loolo

hi

I need your help to resolve this question

y''-y'-2y=18e^-t sin3t

y(c)=0 , y'(c)=3

?

Recall that the laplace transform of

$\mathcal{L}\{e^{at}f(t)\}=F(s-a)$

So the laplace transform of

$f(t)=18e^{-t}\sin(3t) \to \frac{54}{(s+1)^2+9}$

Can you finish from here?

**loolo** · October 5th 2012, 11:32 AM

ammmmmm

I want a solution by this law

Question on the applications of Laplace transforms-03102012975.jpg

**TheEmptySet** · October 5th 2012, 12:32 PM

Originally Posted by loolo

ammmmmm

I want a solution by this law

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I don't see how that rule applies to this problem. Since we don't know the intial contidtions at zero we assume

$y(0)=c_1 \quad y'(0)=c_2$

If we take the laplace transfrom of the equation we get

$s^2Y-sc_1-c_2-sY+c_1-2Y=\frac{54}{(s+1)^2+9}$

Solving for Y gives

$Y=\frac{c_2-c_1}{(s-2)(s+1)}+\frac{c_1s}{(s-2)(s+1)}+\frac{54}{(s-2)(s+1)[(s+1)^2+9]}$

Edit: You could use the theorem to invert the 2nd of the two terms in this problem.

**loolo** · October 5th 2012, 01:06 PM

i Began resolve the question as follows

Question on the applications of Laplace transforms-04102012976.jpg

**TheEmptySet** · October 5th 2012, 01:46 PM

Originally Posted by loolo

i Began resolve the question as follows

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The laplace transform of the product of two function is NOT the product of their transforms.

To transform the function on the right hand side you must use the s-axis translation theorem. See post #2

**loolo** · October 5th 2012, 02:23 PM

ammmmmm

I'll try to find a way s-axis translation theorem.

**Prove It** · October 5th 2012, 05:01 PM

Are you able to use Partial Fractions to simplify before trying to take the Inverse transform to get back to y? If not, you may need to use the convolution theorem...

**loolo** · October 6th 2012, 01:30 AM

Ammmmmm

Dr. asked us to use partial fractions
But I do not know how

**loolo** · October 6th 2012, 02:05 AM

I'm so sorry,

the Conditions in

y(0)=0 ,

y(0)=3

this is my Try...

Question on the applications of Laplace transforms-04102012984.jpg

**loolo** · October 6th 2012, 03:28 AM

Now I want to use partial fractions..

Can you help me ؟

Question on the applications of Laplace transforms-04102012985.jpg

**Prove It** · October 6th 2012, 05:53 AM

Originally Posted by loolo

Now I want to use partial fractions..

Can you help me ؟

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First of all you have an arithmetic error. When you add 3 to the other side and make a common denominator, it should be $\displaystyle \begin{align*} \frac{3\left[ (s + 1)^2 + 9 \right]}{(s + 1)^2 + 9} \end{align*}$ , which when you expand it out gives $\displaystyle \begin{align*} \frac{3(s + 1)^2 + 27}{(s + 1)^2 + 9} \end{align*}$ . As for the rest, you need to solve for $\displaystyle \begin{align*} Y(s) \end{align*}$ then take the inverse transform to get back to $\displaystyle \begin{align*} y(t) \end{align*}$ .

**loolo** · October 6th 2012, 04:50 PM

ammmmmmm

Can you help me complete the answer

**MaxJasper** · October 6th 2012, 06:52 PM

Originally Posted by loolo

I need your help to resolve this question
y''-y'-2y=18e^-t sin3t
y(c)=0 , y'(c)=3

In the diff eq, change t to t+c and take LaplaceT and solve for y(t).

$\mathcal{L}_t[y(t)](s)$ = $\frac{3 e^{-c} \left(e^c s^2+2 e^c s+6 s \sin (3 c)+10 e^c+6 \sin (3 c)+18 \cos (3 c)\right)}{\left(s^2-s-2\right) \left(s^2+2 s+10\right)}$

and an inverse LapT would look like:

$y(t)=e^{-3 c-t} \left(e^{c+3 t}+e^{3 t} \sin (3 c)-e^{3 c} \sin (3 t)+e^{3 t} \cos (3 c)+e^{3 c} \cos (3 t)-e^{4 c}-2 e^{3 c} \cos (3 c)\right)$

**loolo** · October 6th 2012, 06:58 PM

Originally Posted by MaxJasper

In the diff eq, change t to t+c and take LaplaceT and solve for y(t).

$\mathcal{L}_t[y(t)](s)$ = $\frac{3 e^{-c} \left(e^c s^2+2 e^c s+6 s \sin (3 c)+10 e^c+6 \sin (3 c)+18 \cos (3 c)\right)}{\left(s^2-s-2\right) \left(s^2+2 s+10\right)}$

and an inverse LapT would look like:

$y(t)=e^{-3 c-t} \left(e^{c+3 t}+e^{3 t} \sin (3 c)-e^{3 c} \sin (3 t)+e^{3 t} \cos (3 c)+e^{3 c} \cos (3 t)-e^{4 c}-2 e^{3 c} \cos (3 c)\right)$

Originally Posted by loolo

I'm so sorry,

the Conditions in

y(0)=0 ,

y(0)=3

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