![L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\](/tpl/images/0674/0354/8f608.png)
(this is the algebraic equation)![sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}](/tpl/images/0674/0354/ad8ab.png)
(this is the solution for Y(s))![y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}](/tpl/images/0674/0354/130f8.png)
![y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}](/tpl/images/0674/0354/d96fb.png)
Consider the initial value problem y′+2y=4t, y(0)=8.
a. Take the Laplace transform of...
Mathematics, 02.06.2020 20:57, mattiemattoos
Consider the initial value problem y′+2y=4t, y(0)=8.
a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y(s).
b. Solve your equation for Y(s).
c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t).
Answers: 3
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