Laplace inverse of a differential equation with initial conditions: 2(dy/dt) + y =0 , y(0)=-3

Question

1. The exercise:

2. Apply Laplace:

Is it correct? Because I´m confused in 3 or 3s?

3. Finally this:

In this part how to apply Decomposition Partial Fractions ? And how's continued with de System of Equations ? Because with the valors of the variables I know that I have to remplace and apply inverse Laplace

HELP PLEASE !

Answer 1

$$2y'+y=0 \\ y(0)=-3$$ Apply Laplace 's Transform: $$2(sY(s)-y(0))+Y(s)=0$$ $$2sY(s)-\color {red}{2y(0)}+Y(s)=0$$ $$2sY(s)+6+Y(s)=0$$ $$Y(s)(2s+1)=-6$$ $$Y(s)=-\dfrac 6 {(2s+1)}$$ You made some little mistakes. At point 2. The initial condition is multiply by $-2$. So that $-2y(0)=6$.

Apply inverse Laplace Transform: $$Y(s)=-\dfrac 3 {(s+1/2)} \implies y(t)=- 3 e^{-t/2}$$ Since the inverse Laplace Transform for $\dfrac 1 {s+a}$ is $e^{-at}$

Answer 2

Since $$ {l}\{ y(t) \} = s \, y(s) - y(0)$$ then \begin{align} a \, y'(t) + y(t) &= 0 \\ a( s y(s) - y(0)) + y(s) &= 0 \\ (a s + 1) \, y(s) &= a \, y(0) \\ y(s) &= \frac{a \, y(0)}{a s + 1} = \frac{y(0)}{s + \frac{1}{a}} \\ y(t) &= y(0) \, e^{-t/a}. \end{align}

Applying the condition and appropriate constant yields the result.