Question

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

$\begin{array}{l}\mathbf{2}\mathbf{.}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}\\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{8}\mathbf{x}\mathbf{-}\mathbf{t}\\ \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

Short answer

The solution of the given system of differential equations is,

$\begin{array}{l}x\left(t\right)=\frac{t}{8}-\frac{1}{15}{e}^t+\frac{173}{192}{e}^{4t}+\frac{53}{320}{e}^{-4t}\\ y\left(t\right)=\frac{1}{16}-\frac{8}{15}{e}^t+\frac{173}{96}{e}^{4t}-\frac{53}{160}{e}^{-4t}\end{array}$

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

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,

$\begin{array}{l}\frac{dx}{dt}=2y+e^t\\ x\text{'}-2y=e^t\\ sX\left(s\right)-1-2Y\left(s\right)=\frac{1}{s-1}\\ sX\left(s\right)-2Y\left(s\right)=\frac{1}{s-1}+1..........\left(1\right)\\ And,\\ \frac{dy}{dt}=8x-t\\ y\text{'}-8x=-t\\ sY\left(s\right)-y\left(0\right)-8X\left(s\right)=-\frac{1}{s^2}\\ sY\left(s\right)-8X\left(s\right)=1-\frac{1}{s^2}..........\left(2\right)\end{array}$

Simplify the equations

Now multiply the first equation by 8 and equation (2) with s adding both, we get,

$\begin{array}{l}8\left(sX\right(s)-2Y(s\left)\right)+s\left(sY\right(s)-8X(s\left)\right)=8(\frac{1}{s-1}+1)+s(1-\frac{1}{s^2})\\ -16Y\left(s\right)+s^{2}Y\left(s\right)=8+\frac{8}{s-1}+s-\frac{1}{s}\\ (s+4)(s-4)Y\left(s\right)=\frac{8\left(s\right)(s-1)+8s+s\left(s\right)(s-1)-(s-1)}{s(s-1)(s-4)(s+4)}\\ Y\left(s\right)=\frac{s^3+7s^2-s+1}{s(s-1)(s-4)(s+4)}\\ Theaboveequationcanbewrittenas,\\ Y\left(s\right)=\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s-4}+\frac{D}{s+4}\\ Y\left(s\right)=\frac{A(s-1)(s-4)(s+4)+B\left(s\right)(s-4)(s+4)+C\left(s\right)(s-1)(s+4)+D\left(s\right)(s-1)(s-4)}{s(s-1)(s-4)(s+4)}\\ ComparingboththenumeratorsofY\left(s\right),wehave,\\ s^3+7s^2-s+1=A(s-1)(s-4)(s+4)+B\left(s\right)(s-4)(s+4)+C\left(s\right)(s-1)(s+4)+D\left(s\right)(s-1)(s-4)\end{array}$

Step 4:To find the constant values.

Now, putting S=0, we have the value of A to be,

$\begin{array}{l}{0}^3+7(0{)}^2-0+1=A(0-1)(0-4)(0+4)+B\left(0\right)(0-4)(0+4)+C\left(0\right)(0-1)(0+4)+D\left(0\right)(0-1)(0-4)\\ 1=A\left(16\right)+0+0+0\\ A=\frac{1}{16}\\ Similarly,puttingS=1,wehavethevalueofBtobe,\\ 1^3+7(1{)}^2-1+1=A(1-1)(1-4)(1+4)+B\left(1\right)(1-4)(1+4)+C\left(1\right)(1-1)(1+4)+D\left(1\right)(1-1)(1-4)\\ B=-\frac{8}{15}\\ Similarly,puttingS=4,wehavethevalueofCtobe,\\ 4^3+7(4{)}^2-4+1=A(4-1)(4-4)(4+4)+B\left(4\right)(4-4)(4+4)+C\left(4\right)(4-1)(4+4)+D\left(4\right)(4-1)(4-4)\\ C=\frac{173}{96}\\ Similarly,puttingS=-4,wehavethevalue\mathrm{D}oftobe,\\ -4^3+7(-4{)}^2+4+1=A(-4-1)(-4-4)(-4+4)+B(-4)(-4-4)(-4+4)+C(-4)(-4-1)(-4+4)+D(-4)(-4-1)(-4-4)\\ D=-\frac{53}{160}\\ Therefore,wehavethefunctionas,\\ Y\left(s\right)=\frac{1}{16}\frac{1}{s}-\frac{8}{15}\frac{1}{s-1}+\frac{173}{96}\frac{1}{s-4}-\frac{53}{160}\frac{1}{s+4}\end{array}$

Step 5:To find the solutions.

Now, as we assumed$Y\left(s\right)=L\left\{y\right(t\left)\right\}$, we can thus write that,

role="math" localid="1664183542780" $\begin{array}{l}y\left(t\right)=L^1\left\{Y\right(s\left)\right\}\\ =L^1\{\frac{1}{16}\frac{1}{s}-\frac{8}{15}\frac{1}{s-1}+\frac{173}{96}\frac{1}{s-4}-\frac{53}{160}\frac{1}{s+4}\}\\ =\frac{1}{16}{L}^1\left\{\frac{1}{s}\right\}-\frac{8}{15}{L}^1\left\{\frac{1}{s-1}\right\}+\frac{173}{96}{L}^1\left\{\frac{1}{s-4}\right\}-\frac{53}{160}{L}^1\left\{\frac{1}{s+4}\right\}\\ y\left(t\right)=\frac{1}{16}-\frac{8}{15}{e}^t+\frac{173}{96}{e}^{4t}-\frac{53}{160}{e}^{-4t}\\ Fromthegivensystemofequationswehave,\\ \frac{dy}{dt}=8x-t\\ 8x\left(t\right)=\frac{dy\left(t\right)}{dt}+t\\ x\left(t\right)=\frac{t}{8}+\frac{1}{8}\frac{dy\left(t\right)}{dt}\\ x\left(t\right)=\frac{t}{8}-\frac{1}{15}{e}^t+\frac{173}{192}{e}^{4t}+\frac{53}{320}{e}^{-4t}\\ Result:\\ x\left(t\right)=\frac{t}{8}-\frac{1}{15}{e}^t+\frac{173}{192}{e}^{4t}+\frac{53}{320}{e}^{-4t}\\ y\left(t\right)=\frac{1}{16}-\frac{8}{15}{e}^t+\frac{173}{96}{e}^{4t}-\frac{53}{160}{e}^{-4t}\end{array}$