Question

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

$\begin{array}{l}\mathbf{6}\mathbf{.}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{x}\mathbf{-}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\\ \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

Short answer

The solution of the given system of differential equations is,

Step-by-step solution

Definition of Laplace Transform

Let f 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

If$X\left(s\right)=L\left\{x\right(t\left)\right\}andY\left(s\right)=L\left\{y\right(t\left)\right\}$, then after transforming each of the given equations, we obtain,

$\begin{array}{l}\frac{dx}{dt}+x-\frac{dy}{dt}+y=0\\ x\text{'}+x-y\text{'}+y=0\\ sX\left(s\right)-x\left(0\right)+X\left(s\right)-sY\left(s\right)+y\left(0\right)+Y\left(s\right)=0\\ (s+1)X\left(s\right)-(s-1)Y\left(s\right)=-1..........\left(1\right)\\ And,\\ \frac{dx}{dt}+\frac{dy}{dt}+2y=0\\ x\text{'}+y\text{'}+2y=0\\ sX\left(s\right)-x\left(0\right)+sY\left(s\right)-y\left(0\right)+2Y\left(s\right)=0\\ sX\left(s\right)+(s+2)Y\left(s\right)=1..........\left(2\right)\end{array}$

Simplify the equations

Now multiply the first equation with s and equation (2) with (s+1) adding both

, we get,

$\begin{array}{l}s\left(\right(s+1\left)X\right(s)-(s-1\left)Y\right(s\left)\right)-(s+1)sX\left(s\right)+(s+2)Y\left(s\right)=s(-1)-(s+1)\left(1\right)\\ (2s^2+2s+2)Y\left(s\right)=2s+1\\ 2(s^2+s+1)=2(s+\frac{1}{2})\\ Y\left(s\right)=\frac{s+\frac{1}{2}}{s^2+s+1}\\ Fortheabove,wecanwritethedenominatoras,\\ Y\left(s\right)=\frac{s+1/2}{s^2+s+1+{(1/2)}^2-{(1/2)}^2}\\ Y\left(s\right)=\frac{s+1/2}{{(s+1/2)}^2+{(\sqrt{3}/2)}^2}\\ Now,asweassumedY\left(s\right)=L\left\{y\right(t\left)\right\},wecanthuswritethat,\end{array}$

\(\begin{aligned}{l}y(t) = {L^{ - 1}}\left\{ {Y(s)} \right\}\\ = {L^{ - 1}}\{ dfracs + 1/2{(s + 1/2)^2} + {(\sqrt 3 /2)^2}\end{aligned}\)

So, the Laplace inverse is,

\(y(t) = {e^{ - t/2}} \cdot \cos \frac{{\sqrt 3 }}{2}t\)

Step 4:To find the solutions.

From equation (2), we have

$\begin{array}{l}sX\left(s\right)+(s+2)Y\left(s\right)=1\\ sX\left(s\right)=1-\frac{(s+2)(s+1/2)}{s^2+s+1}\\ sX\left(s\right)=\frac{-(3/2)s}{s^2+s+1}\\ X\left(s\right)=\frac{-(3/2)}{s^2+s+1}\\ Fromtheabove,wecanwritethedenominatoras,\\ X\left(s\right)=\frac{-(3/2)}{{(s+1/2)}^2+{(\sqrt{3}/2)}^2}\end{array}$

Now, as we assumed \(X(s) = L\{ x(t)\} \), we can thus write that,

\(\begin{aligned}{l}x(t) = {L^{ - 1}}\left\{ {X(s)} \right\}\\ = {L^{ - 1}}\left\{ {\frac{{ - (3/2)}}{{{{(s + 1/2)}^2} + {{(\sqrt 3 /2)}^2}}}} \right\}\end{aligned}\)

So, the Laplace inverse is

\(x(t) = - \sqrt 3 {e^{ - t/2}} \cdot \sin \frac{{\sqrt 3 }}{2}t\)

Result:

\(x(t) = - \sqrt 3 {e^{ - t/2}} \cdot \sin \frac{{\sqrt 3 }}{2}t\)

\(y(t) = {e^{ - t/2}} \cdot \cos \frac{{\sqrt 3 }}{2}t\)