# Thread: Question on the applications of Laplace transforms

1. ## Question on the applications of Laplace transforms

hi

I need your help to resolve this question

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

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

?

2. ## Re: Question on the applications of Laplace transforms

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

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

So the laplace transform of

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

Can you finish from here?

3. ## Re: Question on the applications of Laplace transforms

ammmmmm

I want a solution by this law

4. ## Re: Question on the applications of Laplace transforms

Originally Posted by loolo
ammmmmm

I want a solution by this law

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

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

If we take the laplace transfrom of the equation we get

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

Solving for Y gives

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

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

5. ## Re: Question on the applications of Laplace transforms

i Began resolve the question as follows

6. ## Re: Question on the applications of Laplace transforms

Originally Posted by loolo
i Began resolve the question as follows

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

7. ## Re: Question on the applications of Laplace transforms

ammmmmm

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

8. ## Re: Question on the applications of Laplace transforms

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...

9. ## Re: Question on the applications of Laplace transforms

Ammmmmm

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

10. ## Re: Question on the applications of Laplace transforms

I'm so sorry,

the Conditions in

y(0)=0 ,

y(0)=3

this is my Try...

11. ## Re: Question on the applications of Laplace transforms

Now I want to use partial fractions..

Can you help me ؟

12. ## Re: Question on the applications of Laplace transforms

Originally Posted by loolo
Now I want to use partial fractions..

Can you help me ؟

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 \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 \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 \displaystyle \begin{align*} Y(s) \end{align*} then take the inverse transform to get back to \displaystyle \displaystyle \begin{align*} y(t) \end{align*}.

13. ## Re: Question on the applications of Laplace transforms

ammmmmmm

Can you help me complete the answer

14. ## Re: Question on the applications of Laplace transforms

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).

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

and an inverse LapT would look like:

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

15. ## Re: Question on the applications of Laplace transforms

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

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

and an inverse LapT would look like:

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

Originally Posted by loolo
I'm so sorry,

the Conditions in

y(0)=0 ,

y(0)=3

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