Question

In Problems 1–12 use the Laplace transform to solve the given initial value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The solution for the given question is $y\left(t\right)=-\frac{1}{4}-\frac{1}{2}t+\frac{1}{4}{e}^{2t}+\left[e^{(t-2)}\sinh (t-2)\right]u(t-2)+e^t\sinh t$.

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.

Solveusing Laplace Transform

Applying Laplace operator in both sides of equation $y\text{'}\text{'}-2y\text{'}=1+\delta \left(t-2\right)$, 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:

$\mathcal{L}\left\{y\text{'}\text{'}\right\}-2\mathcal{L}\left\{y\text{'}\right\}=\mathcal{L}\left\{1\right\}+\mathcal{L}\left\{\delta \right(t-2\left)\right\}$

Apply the property of Dirac delta function${\int }_0^{\infty }f\left(t\right)\delta \left(t-t_0\right)=f\left(t_0\right)$, to write the Laplace of the above equation as follows:

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

On applying given initial Conditions, we get

$$\begin{gathered} Y\left(s\right)=\frac{1}{s\left(s^2-2s\right)}+\frac{e^{-2s}}{s^2-2s}+\frac{1}{s^2-2s} \\ =\frac{e^{-s}}{{(s-1)}^2-1}+\frac{1}{s^2(s-2)}+\frac{1}{{(s-1)}^2-1} \end{gathered}$$

Make the partial fractions of the middle term as:

$\frac{1}{s^2(s-2)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-2}$

Evaluating constant coefficient, we obtain

.$A=-\frac{1}{4}B=-\frac{1}{2}C=\frac{1}{4}$

Thus$Y\left(s\right)=-\frac{1}{4}\frac{1}{s}-\frac{1}{2}\frac{1}{s^2}+\frac{1}{4}\frac{1}{s-2}+\frac{e^{-2s}}{{(s-1)}^2-1}+\frac{1}{{(s-1)}^2-1}$

Take inverse Laplace Transform

For evaluating y(t), utilize inverse form of second translation theorem as:

${\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\left[-\frac{1}{4}\frac{1}{s}-\frac{1}{2}\frac{1}{s^2}+\frac{1}{4}\frac{1}{s-2}\right]\right\}+{\mathcal{L}}^{-1}\left\{\frac{e^{-2s}}{{(s-1)}^2-1}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{{(s-1)}^2-1}\right\}$

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

.$y\left(t\right)=-\frac{1}{4}-\frac{1}{2}t+\frac{1}{4}{e}^{2t}+\left[e^{(t-2)}\sinh (t-2)\right]u(t-2)+e^t\sinh t$