Question

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

${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{2}$

Short answer

The Laplace transform of the given derivative equation$y^{\text{'}}+6y=e^{4t},y\left(0\right)=2$is$y\left(t\right)=\frac{1}{10}{e}^{4t}+\frac{19}{10}{e}^{-6t}$

Step-by-step solution

Theorem of Transform of a Derivative

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

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

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

The given derivative equation is$y^{\text{'}}+6y=e^{4t},y\left(0\right)=2$

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

$$\begin{gathered} \mathrm{L}\left\{{\mathrm{y}}^{\text{'}}+6\mathrm{y}\right\}={\mathrm{Le}}^{4\mathrm{t}} \\ \mathrm{sY}-2+6\mathrm{Y}=\frac{1}{\mathrm{s}-4}\to \mathrm{Y}=\frac{1}{\left(\mathrm{s}-4\right)\left(\mathrm{s}+6\right)}+\frac{2}{\mathrm{s}+6} \end{gathered}$$

Finding the inverse Transform

Now we need to decompose ,

$$\begin{gathered} \frac{1}{\left(\mathrm{s}-4\right)\left(\mathrm{s}+6\right)}=\frac{\mathrm{a}}{\mathrm{s}-4}+\frac{\mathrm{b}}{\mathrm{s}+6} \\ \frac{1}{\left(\mathrm{s}-4\right)\left(\mathrm{s}+6\right)}=\frac{1}{10\left(\mathrm{s}-4\right)}-\frac{1}{10\left(\mathrm{s}+6\right)} \end{gathered}$$

Now we are going to find the inverse transform,

$$\begin{gathered} \frac{1}{10}{L}^{-1}\frac{1}{s-4}-\frac{1}{10}{L}^{-1}\frac{1}{s-\left(-6\right)}+L^{-1}\frac{2}{s-\left(-6\right)} \\ y\left(t\right)=\frac{1}{10}{e}^{4t}-\frac{1}{10}{e}^{-6t}+2e^{-6t} \\ y\left(t\right)=\frac{1}{10}{e}^{4t}+\frac{19}{10}{e}^{-6t} \end{gathered}$$

Therefore, the solved equation is$\mathrm{y}\left(\mathrm{t}\right)=\frac{1}{10}{\mathrm{e}}^{4\mathrm{t}}+\frac{19}{10}{\mathrm{e}}^{-6\mathrm{t}}$.