How Do You Solve an Initial Value Problem Using Laplace Transforms?

[O.J.]

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!

 

[Rainbow Child]

Let

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

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

\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)

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

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

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

y(t)=\mathcal{L}^{-1}\{F(s)\}

you will arrive at y(t)

 

[O.J.]

I'm arriving at these results:

2s / {(s^2+1)^2(2+1)}, but i can't continue from there... I have an appendix of Laplace transforms in my text but none of them seem to fit this one...

 

[Rainbow Child]

Split the fraction into simpler ones

-\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)}

 

[O.J.]

i tried partial fractions, but how do you assign the contants in the numerator fro such a complex polynomial in the denominator...?

 

[Rainbow Child]

I don't understand the question (my English are pretty poor!)

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

If you mean the term

\frac{1-s}{2\,(1+s^2)}

it's inverse Laplace transform can be evaluated by

\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)

 

[Rainbow Child]

and

\mathcal{L}^{-1}\{\frac{\beta}{(s+\alpha)^2+\beta^2}\}=e^{-\alpha\,t}\sin(\beta\,t)

A good place for a table of these transormations could be

http://www.vibrationdata.com/Laplace.htm

 

[HallsofIvy]

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?