Question

Use the Laplace transform to solve the initial value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{7}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{t}}\mathbf{+}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{+}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{4}\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The obtained solution by solving the initial-value problem using the Laplace transform is$y\left(t\right)=-\frac{1}{25}{e}^{-t}-\frac{1}{5}t{e}^t+\frac{1}{25}{e}^{6t}+\frac{1}{5}{e}^{6(t-2)}u(t-2)-\frac{1}{5}{e}^{(t-2)}u(t-2)+\frac{1}{5}{e}^{6(t-4)}u(t-4)-\frac{1}{5}{e}^{(t-4)}u(t-4)$

Step-by-step solution

Define Laplace Transform

The use of Laplace transformation is to convertdifferential equationsinto algebraic equations. The formula for Laplace transform is $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{·}{\mathit{e}}^{\mathbf{-}\mathbf{s}\mathbf{·}\mathbf{t}}\mathbf{·}\mathit{d}\mathit{t}$,Where, F(s) is the Laplace TransformS is the complex numbert is time and is a real number.

Solve using Laplace Transform

Applying Laplace operator in both sides of equation $y\text{'}\text{'}-7y\text{'}+6y=e^t+\delta \left(t-2\right)+\delta \left(t-4\right)$, using the transform$\mathcal{L}\left\{f^n\left(t\right)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-\dots s{f}^{n-2}\left(0\right)-f^{n-1}\left(0\right)$ as follows:$\mathcal{L}\left\{y\text{'}\text{'}\right\}-7\mathcal{L}\left\{y\text{'}\right\}+6\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{e^t\right\}+\mathcal{L}\left\{\delta \right(t-2\left)\right\}+\mathcal{L}\left\{\delta \right(t-4\left)\right\}$

Apply the property of Dirac delta function,$\mathcal{L}\left\{\delta \left(t-t_0\right)\right\}=e^{-s{t}_0}$and the given initial conditionsto write the Laplace of the above equation as follows:

$$\begin{gathered} \left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-7\left[sY\right(s)-y(0\left)\right]+6Y\left(s\right)=\frac{1}{s-1}+e^{-2s}+e^{-4s} \\ Y\left(s\right)=\frac{1}{(s-1)\left(s^2-7s+6\right)}+\frac{e^{-2s}}{\left(s^2-7s+6\right)}+\frac{e^{-4s}}{\left(s^2-7s+6\right)} \\ Y\left(s\right)=\frac{1}{{(s-1)}^2(s-6)}+\frac{e^{-2s}}{(s-6)(s-1)}+\frac{e^{-4s}}{(s-6)(s-1)} \end{gathered}$$

Make the partial fractions of the first term as:

$$\begin{gathered} \frac{1}{{(s-1)}^2(s-6)}=\frac{A}{(s-1)}+\frac{B}{{(s-1)}^2}+\frac{C}{s-6} \\ \frac{e^{-2s}}{(s-1)(s-6)}=e^{-2s}\left[\frac{A\text{'}}{(s-1)}+\frac{B\text{'}}{s-6}\right] \end{gathered}$$

Evaluating constant coefficient, we obtain

$A=-\frac{1}{25}B=\frac{1}{5}C=\frac{1}{25}andA\text{'}=-\frac{1}{5}B\text{'}=-\frac{1}{25}$

Thus,$Y\left(s\right)=\left[-\frac{1}{25}\frac{1}{s-1}-\frac{1}{5}\frac{1}{{(s-1)}^2}+\frac{1}{25}\frac{1}{s-6}\right]+e^{-2s}\left[-\frac{1}{5}\frac{1}{(s-1)}+\frac{1}{5}\frac{1}{s-6}\right]$

Take inverse Laplace Transform

Use Inverse form of second translation theorem to simplify

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\left[-\frac{1}{25}\frac{1}{s-1}-\frac{1}{5}\frac{1}{{(s-1)}^2}+\frac{1}{25}\frac{1}{s-6}\right]\right\}+{\mathcal{L}}^{-1}\left\{e^{-2s}\left[-\frac{1}{5}\frac{1}{(s-1)}+\frac{1}{5}\frac{1}{s-6}\right]\right\} \\ +{\mathcal{L}}^{-1}\left\{e^{-4s}\left[-\frac{1}{5}\frac{1}{(s-1)}+\frac{1}{5}\frac{1}{s-6}\right]\right\} \end{gathered}$$

Take inverse Laplace using Inverse shift theorem$e^{at}f\left(t\right)={\mathcal{L}}^{-1}\left\{{\left.F\left(s\right)\right|}_{s\to s-a}\right\}$as follows:

$y\left(t\right)=-\frac{1}{25}{e}^{-t}-\frac{1}{5}t{e}^t+\frac{1}{25}{e}^{6t}+\frac{1}{5}{e}^{6(t-2)}u(t-2)-\frac{1}{5}{e}^{(t-2)}u(t-2)+\frac{1}{5}{e}^{6(t-4)}u(t-4)-\frac{1}{5}{e}^{(t-4)}u(t-4)$