LaPlace Transformation DFQ Help

Question

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I am struggling to move beyond the inverse step of taking the LaPlace general solution.

I know we take $$\mathcal{L}\{y''\} + 9\mathcal{L}\{y\} = g(t)$$

I know that $\mathcal{L}\{y''\} = s^2$Y(s) - sf(0) - f'(0)$

So then from simple plug and chug, I get that $\mathcal{L}\{y''\} = s^2Y(s) - 1 - 0.$ This is just directly from the IVP problem.

Likewise I know that $\mathcal{L}\{y'\} = sY(s) - y(0)$.

Combining these together into the given equation $y''+9y'= g(t)$

I solve for $Y(s)$ and get $(g(t)+1)/(s^2+9s).$

Up to this point, I am stuck.

Answer 1

$$y''+9y'=g(t)$$ Take the Laplace transform of both sides $$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=\mathcal {L}(g(t))$$ $$s^2F(s)-1+9sF(s)=\mathcal {L}(g(t))$$ $$F(s)(s^2+9s)=1+\mathcal {L}(g(t))$$ You can use the heaviside function $$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$ $$g(t)=1H(t)+(t-1)(H(t-1))$$ You have also that the laplace transform of $$\mathcal {L}(H(t-a)f(t-a))=e^{-as}F(s)$$

Therefore $$\mathcal {L}(g(t))=\frac 1s+\frac {e^{-s}}{s^2}$$ you have now that $$F(s)(s^2+9s)=1+\frac 1s+\frac {e^{-s}}{s^2}$$ $$F(s)=\frac 1 {s(s+9)}+\frac 1{s^2(s+9)}+\frac {e^{-s}}{s^3(s+9)}$$ You can use fractions decompositions ..

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For the first fraction we decompose this way $$h(s)=\frac 1 {s(s+9)}=\frac 19(\frac 1s-\frac 1{s+9})$$ then with the laplace transform table you see that $$\mathcal{L}(h(s))=\frac 19 (1-e^{-9t})$$

Edit2 for the last fraction dont pay attention to the exponential put it outside $$f(s)=\frac {e^{-s}}{s^3(s+9)}=e^{-s}\left (\frac {1}{s^3(s+9)} \right)$$ then perform fraction decomposition...the exponential is just a shift