Question

Someone tells you that the solutions of the two IVPs

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{,} \\ \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathit{\delta }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{,} \end{gathered}$$$$\begin{gathered} \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

are exactly the same. Do you agree or disagree? Defend your answer

Short answer

The solution of the two IVPs is,

$y\left(t\right)=\frac{1}{3}{e}^{-t}\sin 3t$

Step-by-step solution

Step 1:Definition of Laplace Transform

Let$f$ be a function defined for $t⩾0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of$f$ , provided that the integral converges.

Step 2:Use Laplace Transform

We know that

$\mathcal{L}\left\{x^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{x\right\}-sx\left(0\right)-x^{\text{'}}\left(0\right)$

$\mathcal{L}\left\{x^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)$

$\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)$

$\mathcal{L}\left\{y^{\text{'}}\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)$

Take the Laplace transform on both sides of each given differential equation

$\mathcal{L}\left\{0\right\}=\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}+2\mathcal{L}\left\{y\text{'}\right\}+10\mathcal{L}\left\{y\right\}$------------ (1)

$\mathcal{L}\left\{\delta \left(t\right)\right\}=\mathcal{L}\left\{y^{\text{'}\text{'}}\right\}+2\mathcal{L}\left\{y\text{'}\right\}+10\mathcal{L}\left\{y\right\}$------------ (2)

Substitute the known values,

$\mathcal{L}\left\{0\right\}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+2(s\mathcal{L}\{y\}-y(0\left)\right)+10\mathcal{L}\left\{y\right\}$

$\mathcal{L}\left\{\delta \left(t\right)\right\}=s^2\mathcal{L}\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+2(s\mathcal{L}\{y\}-y(0\left)\right)+10\mathcal{L}\left\{y\right\}$

$0=s^2\mathcal{L}\left\{y\right\}-1+2s\mathcal{L}\left\{y\right\}+10\mathcal{L}\left\{y\right\}$$\left(y\left(0\right)=0,y^{\text{'}}\left(0\right)=1\right)$

$\mathcal{L}\left\{\delta \left(t\right)\right\}=s^2\mathcal{L}\left\{y\right\}+2s\mathcal{L}\left\{y\right\}+10\mathcal{L}\left\{y\right\}$$\left(y\left(0\right)=0,y^{\text{'}}\left(0\right)=0\right)$

Denote$\mathcal{L}\left\{y\right\}=Y\left(s\right)$in the above system,

$$\begin{gathered} 1=s^{2}Y\left(s\right)+2sY\left(s\right)+10Y\left(s\right) \\ \mathcal{L}\left\{\delta \left(t\right)\right\}=s^{2}Y\left(s\right)+2sY\left(s\right)+10Y\left(s\right) \end{gathered}$$

Since,$\mathcal{L}\left\{\delta \left(t\right)\right\}=1$,so

$$\begin{gathered} 1=s^{2}Y\left(s\right)+2sY\left(s\right)+10Y\left(s\right) \\ 1=s^{2}Y\left(s\right)+2sY\left(s\right)+10Y\left(s\right) \end{gathered}$$

Simplify the equations

System simplifies to the form.

$$\begin{gathered} \left(s^2+2s+10\right)Y\left(s\right)=1 \\ \left(s^2+2s+10\right)Y\left(s\right)=1 \end{gathered}$$

Note that the equations are the same, so both of the initial value problems will have the same solution.

$Y\left(s\right)=\frac{1}{\left(s^2+2s+10\right)}$

Step 4:Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

${\mathcal{L}}^{-1}\left\{Y\right(s\left)\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{\left(s^2+2s+10\right)}\right\}$

$$\begin{gathered} ={\mathcal{L}}^{-1}\left\{\frac{1}{(s^2+2s+9+1}\right\} \\ =\frac{1}{3}{\mathcal{L}}^{-1}\left\{\frac{3}{{(s+1)}^2+3^3}\right\} \\ =\frac{1}{3}{e}^{-t}\sin 3t \end{gathered}$$

Therefore,

$y\left(t\right)=\frac{1}{3}{e}^{-t}\sin 3t$