Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

 ${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{2}$$\mathit{x}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$

 ${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{x}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{t}\mathbf{-}\mathbf{1}$$\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The solution is  $x\left(t\right)=t^2,y\left(t\right)=t-1$.

 

Step 1: Given information

 Since the initial conditions are given at $t=1$, we need to shift the argument to the zero, so let

 $\begin{array}{l}{x}_1\left(t\right)=x(t+1)\\ y_1\left(t\right)=y(t+1)\end{array}$

Then $x_1\left(0\right)=x\left(1\right)=1$ and $y_1\left(0\right)=y\left(1\right)=0$ and

 $\begin{array}{l}{x}_1^{\text{'}}\left(t\right)=x^{\text{'}}(t+1)(t+1)=x^{\text{'}}(t+1)\\ y_1^{\text{'}}\left(t\right)=y^{\text{'}}(t+1)(t+1)=y^{\text{'}}(t+1)\end{array}$

Then the given system becomes

$\{\begin{array}{l}{x}_1^{\text{'}}-2y_1=2\\ x_1^{\text{'}}+x_1-y_1^{\text{'}}={(t+1)}^2-2(t+1)-1\end{array}\Rightarrow \{\begin{array}{l}{x}_1^{\text{'}}-2y_1=2,x_1\left(0\right)=1\\ x_1^{\text{'}}+x_1-y_1^{\text{'}}=t^2+4t+2,y_1\left(0\right)=0\end{array}$

 

Step 2: Take Laplace transform

Applying the Laplace transform gives

 $\{\begin{array}{l}L\{x_1^{\text{'}}-2y_1\}\left(s\right)=L\left\{2\right\}\left(s\right)\\ L\{x_1^{\text{'}}+x_1-y_1^{\text{'}}\}\left(s\right)=L\{t^2+4t+2\}\left(s\right)\end{array}$

Simplify equation , further as:

 $\Rightarrow \{\begin{array}{l}s{X}_1\left(s\right)-x_1\left(0\right)-2Y_1\left(s\right)=\frac{2}{s}\\ s{X}_1\left(s\right)-x_1\left(0\right)+X_1\left(s\right)-s{Y}_1\left(s\right)-y_1\left(0\right)=\frac{2}{s^3}+\frac{4}{s^2}+\frac{2}{s}\end{array}$

$\Rightarrow \{\begin{array}{l}s{X}_1\left(s\right)-1+2Y_1\left(s\right)=\frac{2}{s}\\ s{X}_1\left(s\right)-1+X_1\left(s\right)-s{Y}_1\left(s\right)=\frac{2}{s^3}+\frac{4}{s^2}+\frac{2}{s}\end{array}$

$\Rightarrow \{\begin{array}{l}{X}_1\left(s\right)=\frac{2}{s^2}+\frac{2}{s}{Y}_1\left(s\right)+\frac{1}{s}\\ (2+\frac{2}{s}-s)Y_1\left(s\right)=\frac{2}{s^3}-\frac{1}{s}+\frac{2}{s^2}\end{array}$

Simplify equation$(2+\frac{2}{s}-s)Y_1\left(s\right)=\frac{2}{s^3}-\frac{1}{s}+\frac{2}{s^2}$ , further as:

 $\begin{array}{c}{Y}_1\left(s\right)=\frac{2-s^2+2s}{s^3}\\ =\frac{1}{s^2}\end{array}$

Step 3: Finding the functions

Substituting  $\frac{1}{s^2}$ for   $Y_1\left(s\right)$ in equation (1) gives

 $X_1\left(s\right)=\frac{2}{s^2}+\frac{2}{s^3}+\frac{1}{s}$

Applying inverse Laplace we get

 $\begin{array}{l}{L}^{-1}\left\{Y_1\left(s\right)\right\}\left(t\right)=L^{-1}\left\{\frac{1}{s^2}\right\}\left(t\right)Þy_1\left(t\right)=t\\ L^{-1}\left\{X_1\left(s\right)\right\}\left(t\right)=L^{-1}\{\frac{2}{s^2}+\frac{2}{s^3}+\frac{1}{s}\}\left(t\right)\end{array}$

Take inverse Laplace transform we get;

 $\begin{array}{c}{x}_1\left(t\right)=t^2+2t+1\\ ={\left(t+1\right)}^2\end{array}$

Now we find

 $\begin{array}{l}x\left(t\right)=x_1(t-1)=t^2\\ y\left(t\right)=y_1(t-1)=t-1\end{array}$

Therefore

 $x\left(t\right)=t^2,y\left(t\right)=t-1$