Question

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

${\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{6}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

The Laplace transform of the given derivative equation ${\mathrm{y}}^{\text{'}\text{'}}-4{\mathrm{y}}^{\text{'}}=6{\mathrm{e}}^{3\mathrm{t}}-3{\mathrm{e}}^{-\mathrm{t}},\mathrm{y}\left(0\right)=1,{\mathrm{y}}^{\text{'}}\left(0\right)=-1$is$\mathrm{Y}\left(\mathrm{t}\right)=\frac{5}{2}+\frac{11}{10}{\mathrm{e}}^{4\mathrm{t}}-2{\mathrm{e}}^{3\mathrm{t}}-\frac{3}{5}{\mathrm{e}}^{-\mathrm{t}}$

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${\mathbf{f}}^{\left(n\right)}\left(t\right)$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$y^{\text{'}\text{'}}-4y^{\text{'}}=6e^{3t}-3e^{-t},y\left(0\right)=1,y^{\text{'}}\left(0\right)=-1$

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

$$\begin{gathered} \mathrm{L}\left\{{\mathrm{y}}^{\text{'}\text{'}}-4{\mathrm{y}}^{\text{'}}\right\}=\mathrm{L}\left\{6{\mathrm{e}}^{3\mathrm{t}}-3{\mathrm{e}}^{-\mathrm{t}}\right\} \\ {\mathrm{s}}^2\mathrm{Y}-\mathrm{s}\left(1\right)+1+-4\mathrm{sY}+4=\frac{6}{\mathrm{s}-3}-\frac{3}{\mathrm{s}+1}\to \mathrm{Y}=\frac{6}{\left(\mathrm{s}-3\right)\left({\mathrm{s}}^2-4\mathrm{s}\right)}-\frac{3}{\left(\mathrm{s}+1\right)\left({\mathrm{s}}^2-4\mathrm{s}\right)}+\frac{\mathrm{s}-5}{{\mathrm{s}}^2-4\mathrm{s}} \end{gathered}$$

Finding the inverse Transform

Now we need to decompose,

$$\begin{gathered} \frac{6}{\left(\mathrm{s}-3\right)\left({\mathrm{s}}^2-4\mathrm{s}\right)}=\frac{\mathrm{a}}{\mathrm{s}-3}+\frac{\mathrm{b}}{\mathrm{s}}+\frac{\mathrm{c}}{\mathrm{s}-4}=\frac{1}{2\mathrm{s}}+\frac{3}{2\left(\mathrm{s}-4\right)}-\frac{2}{\mathrm{s}-3} \\ \frac{3}{\left(\mathrm{s}+1\right)\left({\mathrm{s}}^2-4\mathrm{s}\right)}=\frac{\mathrm{a}}{\mathrm{s}+1}+\frac{\mathrm{b}}{\mathrm{s}}+\frac{\mathrm{c}}{\mathrm{s}-4}=\frac{3}{5\left(\mathrm{s}+1\right)}-\frac{3}{4\mathrm{s}}+\frac{3}{20\left(\mathrm{s}-4\right)} \\ \frac{\mathrm{s}-5}{{\mathrm{s}}^2-4\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{s}}+\frac{\mathrm{b}}{\mathrm{s}-4}=\frac{5}{4\mathrm{s}}-\frac{1}{4\left(\mathrm{s}-4\right)} \end{gathered}$$

Now we are going to find the inverse transform,

$$\begin{gathered} \frac{1}{2}{L}^{-1}\frac{1}{s}+\frac{3}{2}{L}^{-1}\frac{1}{s-4}-2L^{-1}\frac{1}{s-3}- \\ \frac{3}{5}{L}^{-1}\frac{1}{s+1}+\frac{3}{4}{L}^{-1}\frac{1}{s}-\frac{3}{20}{L}^{-1}\frac{1}{s-4}+ \\ \frac{5}{4}{L}^{-1}\frac{1}{s}-\frac{1}{4}{L}^{-1}\frac{1}{s-4} \\ y\left(t\right)=\frac{1}{2}+\frac{3}{2}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t}+\frac{3}{4}-\frac{3}{20}{e}^{4t}+\frac{5}{4}-\frac{1}{4}{e}^{4t} \\ y\left(t\right)=\frac{5}{2}+\frac{11}{10}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t} \end{gathered}$$

Therefore, The Laplace transform of the given derivative equation ${\mathrm{y}}^{\text{'}\text{'}}-4{\mathrm{y}}^{\text{'}}=6{\mathrm{e}}^{3\mathrm{t}}-3{\mathrm{e}}^{-\mathrm{t}},\mathrm{y}\left(0\right)=1,{\mathrm{y}}^{\text{'}}\left(0\right)=-1$is$Y\left(t\right)=\frac{5}{2}+\frac{11}{10}{e}^{4t}-2e^{3t}-\frac{3}{5}{e}^{-t}$