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ggnore's question at Yahoo! Answers (IVP, Laplace transform)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello ggnore,

Using the linearity of $\mathcal{L}$:

$$\mathcal{L}\left\{y''(t)\right\}+8\mathcal{L} \left\{y'(t)\right\}+16\mathcal{L}\left\{y(t) \right\}=0$$ Using $\mathcal{L} \left\{y''(t) \right\}=s^2\mathcal{L} \left\{y(t) \right\}-sy(0)-y'(0)$ and $\mathcal{L}\left\{y'(t)\right\}=s\mathcal{L}\left\{y(t)\right\}-y(0)$ we easily verify: $$\mathcal{L}\left\{y(t)\right\}=\dfrac{s+13}{(s+4)^2}=\ldots=\dfrac{1}{s+4}+\dfrac{9}{(s+4)^2}=Y(s)$$ Using $\mathcal{L}\left\{t^ne^{at}\right\}=\dfrac{n!}{(s-a)^{n+1}}\quad(n=0,1,2,\ldots)$ we get: $$y(t)=\mathcal{L}^{-1}\left\{Y(s)\right\}=e^{-4t}+9te^{-4t}$$