Chapter 9.11, Problem 1AP

Additional Problems

Verify that $x_1\left(t\right)=\left[\begin{array}{c}{e}^{t^2-t}\\ -1\end{array}\right]$ and $x_2\left(t\right)=\left[\begin{array}{c}0\\ 2e^t\end{array}\right]$ are linearly independent solutions to $x^{\prime }=Ax$, where

$A=\left[\begin{array}{cc}2t-1& 0\\ e^{t-t^2}& 1\end{array}\right]$.

Write the general solution to the system $x^{\prime }=Ax$.

To determine

To verify:

That $x_1\left(t\right)$ and $x_2\left(t\right)$ are linearly independent solutions to $x^{\prime }=Ax$.

Answer to Problem 1AP

Solution:

The general solution of the system is $x\left(t\right)=\left[\begin{array}{c}{c}_1{e}^{t^2-t}\\ -c_1+2c_2{e}^t\end{array}\right]$.

Explanation of Solution

Given:

The solutions are given as,

$x_1\left(t\right)=\left[\begin{array}{c}{e}^{t^2-t}\\ -1\end{array}\right]$, $x_2\left(t\right)=\left[\begin{array}{c}0\\ 2e^t\end{array}\right]$ and $A=\left[\begin{array}{cc}2t-1& 0\\ e^{t-t^2}& 1\end{array}\right]$.

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

$x\left(t\right)=c_1{x}_1\left(t\right)+c_2{x}_2\left(t\right)$.

Calculation:

Differentiate the given functions with respect to $t$.

${x^{\prime }}_1\left(t\right)=\left[\begin{array}{c}{e}^{t^2-t}\left(2t-1\right)\\ 0\end{array}\right]$ and ${x^{\prime }}_2\left(t\right)=\left[\begin{array}{c}0\\ 2e^t\end{array}\right]$

Find $A{x}_1$.

$\begin{array}{rll}A{x}_1& =& \left[\begin{array}{cc}2t-1& 0\\ e^{t-t^2}& 1\end{array}\right]\left[\begin{array}{c}{e}^{t^2-t}\\ -1\end{array}\right]\\ & =& \left[\begin{array}{c}{e}^{t^2-t}\left(2t-1\right)\\ 0\end{array}\right]\end{array}$

Find $A{x}_2$.

$\begin{array}{rll}A{x}_2& =& \left[\begin{array}{cc}2t-1& 0\\ e^{t-t^2}& 1\end{array}\right]\left[\begin{array}{c}0\\ 2e^t\end{array}\right]\\ & =& \left[\begin{array}{c}0\\ 2e^t\end{array}\right]\end{array}$

Since, ${x^{\prime }}_1=A{x}_1$ and ${x^{\prime }}_2=A{x}_2$, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

$\begin{array}{rll}W\left[x_1,x_2\right]\left(t\right)& =& \left|\begin{array}{cc}{e}^{t^2-t}& 0\\ -1& 2e^t\end{array}\right|\\ & =& 2e^{t^2}\end{array}$

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

$\begin{array}{rll}x\left(t\right)& =& c_1{x}_1\left(t\right)+c_2{x}_2\left(t\right)\\ & =& c_1\left[\begin{array}{c}{e}^{t^2-t}\\ -1\end{array}\right]+c_2\left[\begin{array}{c}0\\ 2e^t\end{array}\right]\\ & =& \left[\begin{array}{c}{c}_1{e}^{t^2-t}\\ -c_1+2c_2{e}^t\end{array}\right]\end{array}$

Therefore, the general solution of the system is $x\left(t\right)=\left[\begin{array}{c}{c}_1{e}^{t^2-t}\\ -c_1+2c_2{e}^t\end{array}\right]$.

Conclusion:

Hence, the general solution of the system is $x\left(t\right)=\left[\begin{array}{c}{c}_1{e}^{t^2-t}\\ -c_1+2c_2{e}^t\end{array}\right]$.