In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

$\begin{array}{l}{\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{7}\mathbf{,}\\ \mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{9}\end{array}$

The differential equation are given as:

$\begin{array}{l}{x}^{\text{'}\text{'}}+2y^{\text{'}}=-x;x\left(0\right)=2,x^{\text{'}}\left(0\right)=-7,\\ -3x^{\text{'}\text{'}}+2y^{\text{'}\text{'}}=3x-4y;y\left(0\right)=4,y^{\text{'}}\left(0\right)=-9\end{array}$

Given initial value equations are,

$\begin{array}{l}{x}^{\text{'}\text{'}}+2y^{\text{'}}=-x;x\left(0\right)=2,x^{\text{'}}\left(0\right)=-\mathrm{7....}\left(\text{1}\right)\\ -3x^{\text{'}\text{'}}+2y^{\text{'}\text{'}}=3x-4y;y\left(0\right)=4,y^{\text{'}}\left(0\right)=-\mathrm{9...}\left(\text{2}\right)\end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{c}{s}^{2}x\left(s\right)-sx\left(0\right)-x^{\text{'}}\left(0\right)+2\left[sy\right(s)-y(0\left)\right]=-x\left(s\right)\\ s^{2}x\left(s\right)-2s+7+2\left[sy\right(s)-4]=-x\left(s\right)\\ (s^2+1)x\left(s\right)+2sy\left(s\right)=2s+\mathrm{1....}\left(3\right)\end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{l}-3\left[s^{2}x\left(s\right)-x^{\text{'}}\left(0\right)\right]+2\left[s^{2}y\left(s\right)-sy\left(0\right)-y^{\text{'}}\left(0\right)\right]=3x\left(s\right)-4y\left(s\right)\\ -3\left[s^{2}x\left(s\right)-2s+7\right]+2\mid s^{2}y\left(s\right)-4s+9]=3x\left(s\right)-4y\left(s\right)\\ -3s^{2}x\left(s\right)+6s-21+2s^{2}y\left(s\right)-8s+18=3x\left(s\right)-4y\left(s\right)\\ 3(s^2+1)x\left(s\right)-2(2+s^2)y\left(s\right)=-3-2s\mathrm{....}\left(4\right)\end{array}$

Solving equation third and fourth we get

$y\left(s\right)=\frac{4s+3}{s^2+3s+2}$

Using partial fraction we can write as

$y\left(s\right)=\frac{5}{s+2}-\frac{1}{s+1}$

Taking inverse Laplace transform we get

$y\left(t\right)=5e^{-2t}-e^{-t}$

$x\left(s\right)=\frac{2s^3-s^2+s+2}{s^4+3s^3+3s^2+3s+2}$

Using partial fraction we can write as

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

Taking inverse Laplace transform we get

$x\left(t\right)=4e^{-2t}-e^{-t}-cost$

Hence

$\begin{array}{l}x\left(t\right)=4e^{-2t}-e^{-t}-cost,y\left(t\right)=5e^{-2t}-e^{-t}\\ \end{array}$

The final solution is

$\begin{array}{l}x\left(t\right)=4e^{-2t}-e^{-t}-cost,y\left(t\right)=5e^{-2t}-e^{-t}\\ \end{array}$