Laplace initial value problem HELP! PLEASE!

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

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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 $$y(t)$$

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...

Split the fraction into simpler ones

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

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

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

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

HallsofIvy