Question

Solve the following sets of equations by the Laplace transform method

$$\begin{gathered} \mathit{z}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0} \\ {\mathit{y}}_{\mathbf{0}}\mathbf{=}{\mathit{z}}_{\mathbf{0}}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{z}\mathbf{=}\mathbf{2} \end{gathered}$$

Short answer

The value of given pair of linear equation is $y=\sin 2t$and$z=\cos 2t-1$.

Step-by-step solution

Given information from question

The sets of equation are $z+2y=0$

$$\begin{gathered} y_0=z_0=0 \\ y\text{'}-2z=2 \end{gathered}$$

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 given pair of linear equation is

$$\begin{gathered} z\text{'}+2y=0 \\ y\text{'}-2z=2 \end{gathered}$$

The initial conditions are$y_0=0$, and$z_0=0$.

Take Laplace transform on both equations

$$\begin{gathered} z\text{'}+2y=0 \\ L\left\{z\text{'}\right\}+2L\left\{y\right\}=0 \end{gathered}$$

……. (1)

Similarly,

$$\begin{gathered} y\text{'}-2z=2 \\ L\left\{y\right\}-2L\left\{z\right\}=L\left\{2\right\} \end{gathered}$$

……….. (2)

Use L35 Laplace transform

$L35$Laplace transform

localid="1659409372775" $$\begin{gathered} L\left\{y\text{'}\right\}=pL\left\{y\right\}-y_0 \\ L\left\{z\text{'}\right\}=2L\left\{y\right\}=0 \\ pL\left\{z\right\}-z_0+2L\left\{y\right\}=0 \end{gathered}$$ ……….. (3)

Solve further

$$\begin{gathered} pZ-0+2Y=0 \\ pZ+2Y=0 \\ L\left\{y\text{'}\right\}-2L\left\{z\right\}=L\left(2\right) \\ pL\left\{y\right\}-y_0-2L\left\{z\right\}=2.\frac{1}{p} \\ pY-0-2Z=\frac{2}{p} \\ pY-2Z=\frac{2}{p} \end{gathered}$$……… (4)

Multiply equation (3) by (2)

$$\begin{gathered} pY+2Z=0 \\ 2pY+4Y=0 \end{gathered}$$

Multiply equation (4) by p

$$\begin{gathered} pY-2Z=\frac{2}{p} \\ {p}^{2}Y-2pZ=\frac{2P}{p} \\ {p}^{2}Y-2pZ=2 \end{gathered}$$

Add equation (4) and (3)

$$\begin{gathered} 2pZ+4Y+p^{2}Y-2pZ=0+2 \\ 4Y+p^{2}Y=2 \\ \left(4+p^2\right)Y=2 \\ Y=\frac{2}{\left(4+p^2\right)} \end{gathered}$$

Solve further

$Y=\frac{2}{\left(4+p^2\right)}$

Apply inverse Laplace transform

Use inverse Laplace transform

$$\begin{gathered} L^{-1}\left(\sin at\right)=\frac{a}{\left(p^2+a^2\right)} \\ Y=\frac{a}{\left(2^2+p^2\right)} \\ Y=L^{-1}\left[\frac{a}{\left(2^2+p^2\right)}\right] \\ Y=\sin 2t \end{gathered}$$

Substitute$\frac{2}{\left(2^2+p^2\right)}$forin equation (3)

$$\begin{gathered} pZ+2Y=0 \\ pZ+2\left(\frac{2}{{\left(-2+\right)}^2}\right)=0 \end{gathered}$$

…….. (5)

Solve further

$pZ=\frac{-4}{\left(p^2+2^2\right)}$

Solve further by partial fraction

Use partial fraction $\frac{-4}{p\left(p^2+4\right)}$

$$\begin{gathered} \frac{-4}{p\left(p^2+4\right)}=\frac{A}{p}+\frac{Bp+c}{p\left(p^2+4\right)} \\ -4=A\left(p^2+4\right)+\left(Bp+c\right)p \\ -4=A{p}^2+4A+B{p}^2+Cp \end{gathered}$$

Compare coefficients

$$\begin{gathered} 4A=-4 \\ A=-1 \\ C=0 \\ A+B=0 \end{gathered}$$

Solve for B

$$\begin{gathered} -1+B=0 \\ B=1 \end{gathered}$$

Further solve

$$\begin{gathered} \frac{-4}{p\left(p^2+4\right)}=\frac{A}{p}+\frac{Bp+C}{\left(p^2+4\right)} \\ \frac{-4}{p\left(p^2+4\right)}=\frac{-1}{p}+\frac{1\left(p\right)+0}{\left(p^2+4\right)} \\ \frac{-4}{p\left(p^2+4\right)}=\frac{-1}{p}+\frac{p}{\left(p^2+4\right)} \end{gathered}$$

Use above relation in equation (5)

$$\begin{gathered} Z=\frac{-4}{p\left(p^2+2^2\right)} \\ Z=\frac{-1}{p}+\frac{p}{\left(p^2+2^2\right)} \end{gathered}$$

Use inverse Laplace transform$L^{-1}\left(\cos at\right)=\frac{p}{\left(p^2+a^2\right)}$and$L^{-1}\left(\frac{1}{p}\right)=1$

$$\begin{gathered} Z=\frac{-1}{p}+\frac{p}{\left(p^2+2^2\right)} \\ z=L^{-1}\left(\frac{p}{\left(p^2+2^2\right)}\right)-L^{-1}\left(\frac{1}{p}\right) \\ z=\cos 2t-1 \end{gathered}$$

Thus, the value of given pair of linear equation is $y=\sin 2t$and$z=\cos 2t-1$