How to solve $ y''+ 5y' + 4y = 0 $ with Laplace transform.

Question

Laplace inverse of a differential equation with initial conditions: $$y'' + 5y' + 4y = 0 \\ y(0)=1, y´(0)=0$$

I need the MAXIMUM DETAIL of this exercise using Laplace inverse, please enter image description here

Answer 1

Use the table for Laplace transforms (http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx). We take the Laplace transform of the equation: $$\begin{aligned} \mathcal{L}\{y''\} +5\mathcal{L}\{y'\} +4\mathcal{L}\{y\} &= 0\\ s^{2}\mathcal{L}\{y\}-s+5s\mathcal{L}\{y\}-5+4\mathcal{L}\{y\} &=0 \end{aligned}$$ Since, according to the table, $\mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0)$, $\mathcal{L}\{y'\}=sY(s)-y(0)$, and we have the initial conditions that $y(0)=1, y'(0)=0$. Gather appropriately: $$ \mathcal{L}\{y\} \left(s^2+5s+4\right)=s+5$$ $$ \implies y = \mathcal{L}^{-1}\{\frac{s+5}{(s+4)(s+1)}\}$$ Now we have to use partial fraction decomposition, $$ \frac{A}{s+4}+\frac{B}{s+1}= \frac{s+5}{(s+4)(s+1)}$$ Multiplying through by the denominator on the right side of the equation to solve for $A$ and $B$, we arrive at: $$ A(s+1)+B(s+4) = s+5$$ In order to solve for $A$ we set $s=-4$, so that $B$ is multiplied by zero, then we have that $-3A=1 \implies A=\frac{-1}{3}$, similarly if $s=-1 \implies B = \frac{4}{3}$. Substituting those values in: $$ y = \mathcal{L}^{-1}\{ \frac{\frac{-1}{3}}{s+4}\}+\mathcal{L}^{-1}\{ \frac{\frac{4}{3}}{s+1}\} $$ Looking at the Laplace transform table, we see that after factoring out the numerator of each fraction, we are left taking the inverse Laplace transform of an expression of the form $\frac{1}{s-(-a)}$, which has inverse Laplace transform of $e^{at}$. Thus we arrive at: $$ y = \frac{-1}{3}e^{-4t}+\frac{4}{3}e^{-t}$$ I hope this provides as much detail as desired!