Q20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the method of Laplace transforms to solve

$\begin{array}{ll}{\mathbf{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}\mathbf{;}& \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\begin{array}{cc}\mathbf{,}& {\mathbf{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\end{array}\\ \mathbf{4}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{6}\mathbf{;}& \mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{.}\end{array}$

Step-by-Step Solution

Answer

 Therefore the solution is

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

Step 1: Given Information

The given differential equations are:

 $\begin{array}{ll}{x}^{\text{'}\text{'}}+y^{\text{'}}=2;& x\left(0\right)=3\begin{array}{cc},& x^{\text{'}}\left(0\right)=0\end{array}\\ 4x+y^{\text{'}}=6;& y\left(1\right)=4.\end{array}$

Step 2: Applying Laplace transform

Using $y\left(\text{0}\right)=c$we get the system

$\{\begin{array}{ll}{x}^{\text{'}\text{'}}+y^{\text{'}}=2,& x\left(0\right)=3,x^{\text{'}}\left(0\right)=0\\ 4x+y^{\text{'}}=6,& y\left(0\right)=c\end{array}$

Applying the Laplace transform we get

$\{\begin{array}{c}\mathcal{L}\{x^{\text{'}\text{'}}+y^{\text{'}}\}=\mathcal{L}\left\{2\right\}\\ \mathcal{L}\{4x+y^{\text{'}}\}=\mathcal{L}\left\{6\right\}\end{array}$

$\{\begin{array}{l}\mathcal{L}\left\{x^{\text{'}\text{'}}\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}=\frac{2}{s}\\ 4\mathcal{L}\left\{x\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}=\frac{6}{s}\end{array}$

$\{\begin{array}{l}{s}^{2}X\left(s\right)-sx\left(0\right)-x^{\text{'}}\left(0\right)+sY\left(s\right)-y\left(0\right)=\frac{2}{s}\\ 4X\left(s\right)+sY\left(s\right)-y\left(0\right)=\frac{6}{s}\end{array}$

Multiplying the second equation with -1 and adding to the first gives

$\begin{array}{c}{s}^{2}X\left(s\right)-4X\left(s\right)-3s=-\frac{4}{s}\\ (s^2-4)X\left(s\right)=3s-\frac{4}{s}\\ (s^2-4)X\left(s\right)=\frac{3}{s}\\ X\left(s\right)=\frac{3s^2-4}{s(s^2-4)}\end{array}$

Decompose the resulting equation as:

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

Step 3: Applying the inverse Laplace

Take inverse Laplace transform on both sides of $X\left(s\right)=\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s-2}$ , as:

$x\left(t\right)=1+e^{-2t}+e^{2t}$

Substituting $\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s-2}$  for$X\left(s\right)$  into the second equation in the system (1) we get

$\begin{array}{c}4(\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s-2})+sY\left(s\right)-c=\frac{6}{s}\\ sY\left(s\right)-c=\frac{6}{s}-\frac{4}{s}-\frac{4}{s+2}-\frac{4}{s-2}\\ Y\left(s\right)=\frac{2}{s^2}-\frac{4}{s(s+2)}-\frac{4}{s(s-2)}+\frac{c}{s}\\ =\frac{2}{s^2}-\frac{2}{s}+\frac{2}{s+2}+\frac{2}{s}-\frac{2}{s-2}+\frac{c}{s}\\ =\frac{2}{s^2}+\frac{2}{s+2}-\frac{2}{s-2}+\frac{c}{s}\end{array}$

Applying the inverse Laplace, we get

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

Using that $y\left(1\right)=4$we obtain

$\begin{array}{c}4=2+2e^{-2}-2e^2+c\\ c=2e^2-2e^{-2}+2\end{array}$

Therefore

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