Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{6}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathbf{}\mathbf{sin}\mathbf{3}\mathit{t}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$

Short answer

The given differential equation's solution is $y\left(t\right)=t{e}^{-t}\cos 3t$.

Step-by-step solution

Given information from question

The differential equation is $y\text{'}\text{'}+2y\text{'}+10y=-6e^{-\mathrm{t}}\sin 3t,y_0=0,y{\text{'}}_0=1$.

Laplace transform

Application of Laplace transform:

It's used to simplify complicated differential equations by using polynomials. It's used to convert derivatives into numerous domain variables, then utilise the Inverse Laplace transform to convert the polynomials back to the differential equation.

Use Laplace transform on both sides of given differential equation

The initial conditions are$y_0=0,y{\text{'}}_0=1$.

Consider L35 result of Laplace transform of derivative of y.

$$\begin{gathered} L\left(y\right)=Y \\ L(y\text{'})=pY-y_0 \\ L(y\text{'}\text{'})=p^{2}Y-p_0-y{\text{'}}_0 \\ L\left(e^{-at}\sin bt\right)=\frac{p+a}{{\left(p+a\right)}^2+b^2},\mathrm{Re}\left(p+a\right)>\left|\mathrm{lm}b\right| \end{gathered}$$

Use Laplace on both sides of given differential equation

$L\left(y\text{'}\text{'}+2y\text{'}+10y\right)=L\left(-6e^{-t}\sin 3t\right)$

Compare right hand side term$L\left(-6e^{-t}\sin 3t\right)$with the general term$L\left(e^{-at}\sin bt\right)$.

$\begin{array}{l}a=1\\ b=3\end{array}$

Apply linearity property of Laplace transformation

Use linearity property of Laplace transformation as

$$\begin{gathered} L\left(y\text{'}\text{'}\right)+2L\left(y\text{'}\right)+10L\left(y\right)=-6L\left(e^{-t}\sin 3t\right) \\ \left(p^{2}Y-p{y}_0-y{\text{'}}_0\right)+4\left(pY-y_0\right)+10y=-6\frac{3}{{\left(p+1\right)}^2+3^2} \\ \left(p^2+2p+10\right)Y=\frac{-18}{{\left(p+1\right)}^2+3^2}+1 \end{gathered}$$

Consider the equation,

$$\begin{gathered} p^2+2p+10=p^2+2p+1+9 \\ ={(p+1)}^2+3^2 \end{gathered}$$

Use this in above equation.

$$\begin{gathered} \left(p^2+2p+10\right)Y=\frac{-18}{{\left(p+1\right)}^2+3^2}+1 \\ Y=\frac{-18}{\left[{\left(p+1\right)}^2+3^2\right]\left[{\left(p+1\right)}^2+3^2\right]}+\frac{1}{{\left(p+1\right)}^2+3^2} \\ y=\frac{-18}{{\left[{\left(p+1\right)}^2+3^2\right]}^2}+\frac{1}{{\left(p+1\right)}^2+3^2} \\ y=\frac{{\left(p+1\right)}^2+3^2}{{\left[{\left(p+1\right)}^2+3^2\right]}^2} \end{gathered}$$

The term $\frac{(p+1{)}^2-3^2}{{\left[{\left(p+1\right)}^2+3^2\right]}^{-2}}$contains variable (p+1) instead of p.

So, use frequency shift theorem to replace p with (p-1). From this replacement, one can use Laplace inverse transformation.

Solve further for frequency shift theorem

The frequency shift theorem

$$\begin{gathered} y\left(t\right)=L^{-1}\left\{Y\left(p\right)\right\} \\ \left\{e^{-p_{0}t}y\left(t\right)\right\}=L^{-1}\left\{Y\left(p+p_0\right)\right\} \end{gathered}$$

Substitute -1 for p and (p - 1) for p in above equation.

Solve further

$e^{2t}y\left(t\right)=L^{-1}\left\{Y\left(p-1\right)\right\}$ ……. (1)

$Y\left(p\right)=\frac{{\left(p+1\right)}^2-3^2}{{\left[{\left(p+1\right)}^2+3^2\right]}^2}$

Substitute p-1 for p in above equation.

$$\begin{gathered} Y\left(p-1\right)=\frac{2\left(p-1+1\right)-3^2}{\left[{\left(p-1+1\right)}^2+3^2\right]} \\ =\frac{p^2-3^2}{{\left(p^2-3^2\right)}^2} \end{gathered}$$

Substitute the value in equation (1)

role="math" localid="1659256010718" $$\begin{gathered} e^{2t}y\left(t\right)=L^{-1}\left\{Y\left(p-1\right)\right\} \\ {e}^{2t}y\left(t\right)=L^{-1}\left[\frac{p^2-3^2}{{\left(p^2+3^2\right)}^2}\right] \\ =t\cos \left(3t\right) \\ t\cos at=L^{-1}\left[\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2}\right] \end{gathered}$$

Divide both side by.

$y\left(t\right)=t{e}^{-t}\cos 3t$

Thus, the given differential equation's solution is $y\left(t\right)=t{e}^{-t}\cos 3t$.