Question

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

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

Short answer

The solution of the given system of differential equations is,

$\begin{array}{l}x\left(t\right)=\frac{1}{3}{e}^t-\frac{1}{3}{e}^{-2t}\\ y\left(t\right)=\frac{2}{3}{e}^t+\frac{1}{3}{e}^{-2t}\end{array}$

Step-by-step solution

Definition of Laplace Transform

Let be a function defined for $t\ge 0$. 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.

Use Laplace Transform

Take the Laplace to transform on both sides of both equations.

Using linearity of the Laplace transform we have:

$\begin{array}{l}x\text{'}=-x+y\to L\left[x\right]-x\left(0\right)=-L\left[x\right]+L\left[y\right]\\ y\text{'}=2x\to sL\left[y\right]-y\left(0\right)=2L\left[x\right]\\ Whenwegroupcoefficientwehave:\\ (s+1)L\left[x\right]-L\left[y\right]=0.......\left(1\right)\\ -2L\left[x\right]+sL\left[y\right]=1........\left(2\right)\end{array}$

Simplify the equations

Now multiply the first equation with s and add it to the second one.

We have:

$\begin{array}{l}s(s+1)L\left[x\right]-2L\left[x\right]=1\\ L\left[x\right](s^2+s-2)=1\\ L\left[x\right]=\frac{1}{s^2+s-2}\\ L\left[x\right]=\frac{1}{(s-1)(s+2)}\end{array}$

Use the technique of partial fractions.

Use the technique of partial fractions.

We have:

$\begin{array}{l}\frac{1}{(s-1)(s+2)}=\frac{A}{(s-1)}+\frac{B}{(s+2)}\\ 1=A(s+2)+B(s-1)\end{array}$

Simplify,

$\begin{array}{l}\to A+B=0\to B=-A\\ \to 2A-B=1\to 2A+A=1\\ A=\frac{1}{3}\\ B=-\frac{1}{3}\end{array}$

So to conclude:

$\begin{array}{l}\frac{A}{(s-1)}+\frac{B}{(s+2)}=\frac{\frac{1}{3}}{(s-1)}+\frac{-\frac{1}{3}}{(s+2)}\\ So,wehave,\\ L\left[x\right]=\frac{\frac{1}{3}}{(s-1)}+\frac{-\frac{1}{3}}{(s+2)}\\ x\left(t\right)=\frac{1}{3}{e}^t-\frac{1}{3}{e}^{-2t}\end{array}$

Step 5:To check the initial values.

We can get y(t) from the first equation of the initial problem.

$\begin{array}{l}y\left(t\right)=x\text{'}\left(t\right)+x\left(t\right)\\ =\frac{1}{3}{e}^t+\frac{2}{3}{e}^{-2t}+\frac{1}{3}{e}^t-\frac{1}{3}{e}^{-2t}\\ =\frac{2}{3}{e}^t+\frac{1}{3}{e}^{-2t}\end{array}$

In this last step, we will check the initial values of each differential equation.

We have:

$\begin{array}{l}y\left(0\right)=1\\ y\left(t\right)=\frac{2}{3}{e}^t+\frac{1}{3}{e}^{-2t}\\ y\left(0\right)=\frac{2}{3}+\frac{1}{3}=1\end{array}$

And,

$\begin{array}{l}x\left(0\right)=0\\ x\left(t\right)=\frac{1}{3}{e}^t-\frac{1}{3}{e}^{-2t}\\ x\left(0\right)=\frac{1}{3}-\frac{1}{3}=0\end{array}$

Result:

localid="1664180838720" $\begin{array}{l}x\left(t\right)=\frac{1}{3}{e}^t-\frac{1}{3}{e}^{-2t}\\ y\left(t\right)=\frac{2}{3}{e}^t+\frac{1}{3}{e}^{-2t}\end{array}$