MHB Ggnore's question at Yahoo Answers (IVP, Laplace transform)

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The discussion addresses an initial value problem defined by the differential equation y'' + 8y' + 16y = 0 with initial conditions y(0) = 1 and y'(0) = 5. The Laplace transform of the solution is derived, resulting in Y(s) = (s + 13) / (s + 4)². The inverse Laplace transform yields the solution function y(t) = e^(-4t) + 9te^(-4t). The response provides a clear method for applying the Laplace transform to solve the given IVP. This approach effectively demonstrates the use of linearity and known transforms in solving differential equations.
Fernando Revilla
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Here is the question:

Consider
y''+8y'+16y=0, y(0)=1, y'(0)=5

find the laplace transform to the solution, that is Y(s)=L(y(t))
Y(s)=_______

Find the function y solution of the IVP abov,
y(t)=__________________

Here is a link to the question:

Consider the initial value problem for y; Laplace? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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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: $$y(t)=\mathcal{L}^{-1}\left\{Y(s)\right\}=e^{-4t}+9te^{-4t}$$
 
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