Q23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

solve the given initial value problem using the method of Laplace transforms. Sketch the graph of the solution.$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{t}\mathbf{-}\mathbf{-}\left(t--4\right)\mathit{u}\left(t--2\right)\mathbf{;}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

On solving the given initial value problem using the method of Laplace transforms the solution is  $y\left(t\right)=t-(t-4+2\cos (t-2)-\sin (t-2\left)\right)u(t-2)$and the graph is

 

Step 1: Definition

The Laplace transform, is an integral transform that converts a function of a real variable usually t, the time domain to a function of a complex variable s. 

                                                                                                      

Step 2: Applying Laplace transform and using its linearity

$$\begin{gathered} \mathcal{L}\left\{y\text{'}\text{'}+y\right\}=\mathcal{L}\{t-(t-4\left)u\right(t-2\left)\right\} \\ \mathcal{L}\left\{y\text{'}\text{'}\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{t\right\}-\mathcal{L}\left\{\right(t-4\left)u\right(t-2\left)\right\} \end{gathered}$$

$$\begin{gathered} \left(s^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right)+Y\left(s\right)=\frac{1}{s^2}-\mathcal{L}\left\{\right(t-4\left)u\right(t-2\left)\right\} \\ {s}^{2}Y\left(s\right)-1+Y\left(s\right)=\frac{1}{s^2}-\mathcal{L}\left\{\right(t-4\left)u\right(t-2\left)\right\} \end{gathered}$$

Since,

 $$\begin{gathered} \mathcal{L}\left\{\right(t-4\left)u\right(t-2\left)\right\}=e^{-2s}\mathcal{L}\{t-2\} \\ =e^{-2s}(\mathcal{L}\{t\}-2\mathcal{L}\{1\left\}\right) \\ =e^{-2s}\left(\frac{1}{s^2}-\frac{2}{s}\right) \end{gathered}$$

We have that,

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

 

Using partial fractions we get,

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

And inverse Laplace transform gives,

 $$\begin{gathered} {\mathcal{L}}^{-1}\left\{\frac{e^{-2s}}{s^2}\right\}=(t-2)u(t-2) \\ {\mathcal{L}}^{-1}\left\{\frac{2e^{-2s}}{s}\right\}=2u(t-2) \end{gathered}$$

$$\begin{gathered} 2{\mathcal{L}}^{-1}\left\{\frac{2e^{-2s}s}{s^2+1}\right\}=2\cos (t-2)u(t-2) \\ {\mathcal{L}}^{-1}\left\{\frac{e^{-2s}}{s^2+1}\right\}=\sin (t-2)u(t-2) \end{gathered}$$

 

Step 3: The solution of given IVP

$$\begin{gathered} y\left(t\right)={\mathcal{L}}^{-1}\left\{\frac{1}{s^2}-e^{-2s}\frac{1-2s}{s^2\left(s^2+1\right)}\right\} \\ ={\mathcal{L}}^{-1}\left\{\frac{1}{s^2}\right\}-{\mathcal{L}}^{-1}\left\{\frac{e^{-2s}}{s^2}-\frac{2e^{-2s}}{s}+\frac{2e^{-2s}s}{s^2+1}-\frac{e^{-2s}}{s^2+1}\right\} \\ =t-\left(\right(t-2\left)u\right(t-2)-2u(t-2)+2\cos (t-2\left)u\right(t-2)-\sin (t-2\left)u\right(t-2\left)\right) \\ =t-(t-4+2\cos (t-2)-\sin (t-2\left)\right)u(t-2) \end{gathered}$$

Hence, $y\left(t\right)=t-(t-4+2\cos (t-2)-\sin (t-2\left)\right)u(t-2)$

 and the graph is