Question

In ProbleIms 1–12 use the Laplace transform to solve the given initial value problem.$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{\delta }\mathbf{(}\mathit{t}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

On applying Laplace the transform for the given initial value, the solution is $y\left(t\right)=e^{-(t-1)}(t-1)u(t-1)$.

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

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

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

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:

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

On applying given initial Conditions, we get

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

Take inverse Laplace Transform

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

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

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:

$y\left(t\right)=e^{-(t-1)}(t-1)u(t-1)$.