Question

In Problems 1-12 Use the Laplace transform to solve the given system of differential equations.

$\begin{array}{l}\mathbf{3}\mathbf{.}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{y}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\end{array}$

Short answer

The solution of the given system of differential equations is,

$\begin{array}{l}x\left(t\right)=-\cos 3t-\frac{5}{3}\sin 3t\\ y\left(t\right)=2\cos 3t-\frac{7}{3}\sin 3t\end{array}$

Step-by-step solution

Definition of Laplace Transform

Let be a function defined for. 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.

Use Laplace Transform

We know that

$\begin{array}{l}\mathcal{L}\left\{x^{\text{'}}\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)...........\left(I\right)\\ \mathcal{L}\left\{y^{\text{'}}\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)............\left(II\right)\\ TaketheLaplacetransformonbothsidesofeachgivendifferentialequation\\ \mathcal{L}\left\{x\right\}-2\mathcal{L}\left\{y\right\}=\mathcal{L}\{x\text{'}\}\\ 5\mathcal{L}\left\{x\right\}-\mathcal{L}\left\{y\right\}=\mathcal{L}\{y\text{'}\}\\ Substitutetheequation\left(1\right)and\left(2\right),weget\\ \mathcal{L}\left\{x\right\}-2\mathcal{L}\left\{y\right\}=s\mathcal{L}\left\{x\right\}-x\left(0\right)\\ 5\mathcal{L}\left\{x\right\}-\mathcal{L}\left\{y\right\}=s\mathcal{L}\left\{y\right\}-y\left(0\right)\\ Wehave,X\left(0\right)=-1,\mathrm{Y}\left(0\right)=2,so\\ \mathcal{L}\left\{x\right\}-2\mathcal{L}\left\{y\right\}=s\mathcal{L}\left\{x\right\}+1\\ 5\mathcal{L}\left\{x\right\}-\mathcal{L}\left\{y\right\}=s\mathcal{L}\left\{y\right\}-2\\ Denote\mathcal{L}\left\{x\right\}=X\left(s\right)and\mathcal{L}\left\{y\right\}=Y\left(s\right)intheabovesystem.\\ X\left(s\right)-2Y\left(s\right)=2X\left(s\right)+1\\ 5X\left(s\right)-Y\left(s\right)=sY\left(s\right)-2\end{array}$

Simplify the equations

The system simplifies the form.

$\begin{array}{l}\left(s-1\right)X\left(s\right)+2Y\left(s\right)=-1-------\left(\mathrm{I}\right)\\ 5X\left(s\right)-\left(s+1\right)Y\left(s\right)=-2---\left(II\right)\\ Multiplyequation\left(3\right)byandequation\left(4\right)by2.\\ \left(s-1\right)(\mathrm{s}+1)X\left(s\right)+2(s+1)Y\left(s\right)=-(s+1)\\ 10X\left(s\right)-2(s+1)Y\left(s\right)=-4\\ Nowadd\left(5\right)and\left(6\right)equationstoget.\\ (s^2-1)X\left(s\right)+10X\left(s\right)=-(s+1)-4\\ \left(s^2-1+10\right)X\left(s\right)=-s-1-4\\ X\left(s\right)=\frac{-s-5}{s^2+9}\\ X\left(s\right)=-\frac{s}{s^2+3^2}-\frac{5}{3}\frac{3}{s^2+3^2}\end{array}$

Use Inverse Laplace Transform

Apply inverse Laplace transform to each side.

Result,

$\begin{array}{l}x\left(t\right)=-\cos 3t-\frac{5}{3}\sin 3t\\ y\left(t\right)=2\cos 3t-\frac{7}{3}\sin 3t\end{array}$