Respuesta :
Answer:
[tex]y(t) = 3x+8e^{-3x} -1[/tex]
Step-by-step explanation:
Recall that the following laplace transforms
[tex]L(y') = sY(s)-y(0)[/tex]
[tex]L(t) = \frac{1}{s^2}[/tex]
The laplace transform is linear, so, applying the laplace transform to the equation we get
[tex]L(y'+3y) = sY(s)-7+3Y(s) = L(9t) = \frac{9}{s^2}[/tex]
By some algebraic manipulations, we get
[tex] Y(s)(s+3) = \frac{9+7s^2}{s^2}[/tex]
which is equivalent to
[tex] Y(s) = \frac{9+7s^2}{s^2(s+3)} = \frac{9}{s^2(s+3)}+\frac{7}{s+3}[/tex]
By using the partial fraction decomposition, we get
[tex] \frac{9}{s^2(s+3)} = \frac{-1}{s} + \frac{3}{s^2} + \frac{1}{s+3}[/tex]
then
[tex]Y(s) = \frac{-1}{s} + \frac{3}{s^2} + \frac{1}{s+3} + \frac{7}{s+3} = \frac{8}{s+3} + \frac{3}{s^2}-\frac{1}{s}[/tex]
Using that
[tex] L(e^{-ax}) = \frac{1}{s+a}[/tex]
[tex]L(1) = \frac{1}{s}[/tex]
by taking the inverse on both sides we get
[tex] y(t) = L^{-1}(\frac{8}{s+3})+L^{-1}(\frac{3}{s^2})+L^{-1}(-\frac{1}{s}) = 8e^{-3x} + 3x-1[/tex]
