Question

Use the Laplace transform to solve the given initial-value problem

${\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{5}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The Laplace transform of the given derivative equation ${\mathrm{y}}^{\text{'}\text{'}}+5{\mathrm{y}}^{\text{'}}+4\mathrm{y}=0,\mathrm{y}\left(0\right)=1,{\mathrm{y}}^{\text{'}}\left(0\right)=0$is$Y\left(t\right)=\frac{4}{3}{e}^{-t}-\frac{1}{3}{e}^{-4t}$

Step-by-step solution

Theorem of Transform of a Derivative

If $\mathbf{f}\mathbf{,}{\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{f}}^{\left(n-1\right)}$are continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and are of exponential order and if${\mathbf{f}}^{\left(n\right)}\left(t\right)$is a piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$, then

$\mathbf{L}\left\{f^{\left(n\right)}\left(t\right)\right\}\mathbf{=}{\mathbf{s}}^{\mathbf{n}}\mathbf{F}\left(s\right)\mathbf{-}{\mathbf{s}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{f}\left(0\right)\mathbf{-}{\mathbf{s}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathbf{f}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{-}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{-}{\mathbf{f}}^{\left(n-1\right)}\left(0\right)$

Where $\mathbf{F}\left(s\right)\mathbf{=}\mathbf{L}\left\{f\left(t\right)\right\}$

The given derivative equation is ${\mathrm{y}}^{\text{'}\text{'}}+5{\mathrm{y}}^{\text{'}}+4\mathrm{y}=0,\mathrm{y}\left(0\right)=1,{\mathrm{y}}^{\text{'}}\left(0\right)=0$

Using the Theorem of Transform of a Derivative, we are transforming the above given derivative,

$$\begin{gathered} L\left\{y^{\text{'}\text{'}}+5y^{\text{'}}+4y\right\}=L\left\{0\right\} \\ {s}^{2}Y-s\left(1\right)-0+5sY-5\left(1\right)+4y=0\to Y=\frac{5+s}{s^2+5s+4}=\frac{5+s}{\left(s+1\right)\left(s+4\right)} \end{gathered}$$

Finding the inverse Transform

Now we need to decompose,

$$\begin{gathered} \frac{5+\mathrm{s}}{\left(\mathrm{s}+1\right)\left(\mathrm{s}+4\right)}=\frac{\mathrm{a}}{\mathrm{s}+1}+\frac{\mathrm{b}}{\mathrm{s}+4} \\ \frac{5+\mathrm{s}}{\left(\mathrm{s}+1\right)\left(\mathrm{s}+4\right)}=\frac{4}{3\left(\mathrm{s}+1\right)}-\frac{1}{3\left(\mathrm{s}+4\right)} \end{gathered}$$

Now we are going to find the inverse transform,

$$\begin{gathered} \frac{4}{3}{L}^{-1}\frac{1}{s-\left(-1\right)}-\frac{1}{3}{L}^{-1}\frac{1}{s-\left(-4\right)} \\ y\left(t\right)=\frac{4}{3}{e}^{-t}-\frac{1}{3}{e}^{-4t} \end{gathered}$$

Therefore, The Laplace transform of the given derivative equation $y^{\text{'}\text{'}}+5y^{\text{'}}+4y=0,y\left(0\right)=1,y^{\text{'}}\left(0\right)=0$is $\mathrm{Y}\left(\mathrm{t}\right)=\frac{4}{3}{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{3}{\mathrm{e}}^{-4\mathrm{t}}$.