Laplace initial value problem HELP! PLEASE!

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  • #1
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Laplace initial value problem.... HELP! PLEASE!

Hello all!
I'm stuck on this question:

y' + y = t sin t

y(0) = 0

solve it using laplace transform,... my final is tomorrow, and its 2 am, i would appreciate a quick respone
thanks in advance!
 
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Answers and Replies

  • #2
Let

[tex]\mathcal{L}\{y(t)\}=F(s)=\int_0^{+\infty} e^{-s\,t}\,y(t)\,d\,t[/tex]

be the Laplace transform of [tex] y(t)[/tex], then the Laplace transform for [tex] \mathcal{L}\{y'(t)\} [/tex] would be

[tex]\mathcal{L}\{y'(t)\}=\int_0^{+\infty} e^{-s\,t}\,y'(t)\,d\,t\Rightarrow \mathcal{L}\{y'(t)\}=s\,F(s)-y(0) \Rightarrow \mathcal{L}\{y'(t)\}=s\,F(s)[/tex]

The Laplace transform of the right hand side of your eq is

[tex] \mathcal{L}\{t\,\sin t\}=-\frac{2\,s}{(1+s^2)^2} [/tex]

Plugging the above values into your equation, you can evaluate [tex]F(s)[/tex] and applying the inverse Laplace transformation

[tex]y(t)=\mathcal{L}^{-1}\{F(s)\}[/tex]

you will arrive at [tex] y(t)[/tex]
 
  • #3
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I'm arriving at these results:

2s / {(s^2+1)^2(2+1)}, but i cant continue from there.... I have an appendix of Laplace transforms in my text but none of them seem to fit this one...
 
  • #4
Split the fraction into simpler ones

[tex]-\frac{2\,s}{(1+s^2)^2\,(1+s)}=\frac{1}{2\,(1+s)}-\frac{1}{(1+s^2)^2}-\frac{s}{(1+s^2)^2}+\frac{1-s}{2\,(1+s^2)}[/tex]
 
  • #5
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i tried partial fractions, but how do you assign the contants in the numerator fro such a complex polynomial in the denominator...?
 
  • #6
I don't understand the question (my English are pretty poor!)

" ...complex polynomial in the denominator..."

If you mean the term

[tex]\frac{1-s}{2\,(1+s^2)}[/tex]

it's inverse Laplace transform can be evaluated by

[tex]\mathcal{L}^{-1}\{\frac{s+\gamma}{(s+\alpha)^2+\beta^2}\}=e^{-\alpha\,t}\left(\cos(\beta t)+\frac{\gamma-\alpha}{\beta}\,\sin(\beta\,t)\right)[/tex]
 
  • #8
HallsofIvy
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The standard technique for finding inverse Laplace transformations of complicated fractions (I wouldn't say "complex"; in mathematics that is too closely connected with complex numbers) is to use partial fractions. Surely you've seen that before?
 

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