Try the fastest way to create flashcards
Question

Show that x1(t) and x2(t)x_{1}(t) \text { and } x_{2}(t) are solutions to the initial-value problem dx/dt=Ax with x(0)=x0.d \mathbf{x} / d t=A \mathbf{x} \text { with } \mathbf{x}(0)=\mathbf{x}_{0}.

A=[1101]x0=[21]A=\left[\begin{array}{rr} {1} & {-1} \\ {0} & {1} \end{array}\right] \quad \mathbf{x}_{0}=\left[\begin{array}{l} {2} \\ {1} \end{array}\right]

x1(t)=2ettet,x2(t)=etx_{1}(t)=2 e^{t}-t e^{t}, x_{2}(t)=e^{t}

Solution

Verified
Answered 1 year ago
Answered 1 year ago
Step 1
1 of 4

Denote:

x=[x1(t)x2(t)]=[2ettetet](1)x=\begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}=\begin{bmatrix} 2e^{t}-te^{t}\\ e^{t} \end{bmatrix}\tag{1}

To prove the functions x1(t)x_{1}(t) and x2(t)x_{2}(t) are solutions to the initial value problem:

dxdt=Ax,\frac{dx}{dt}=Ax,

where

A=[1101],andx0=[21]A=\begin{bmatrix} 1&-1\\ 0&1 \end{bmatrix},\quad\text{and}\quad x_{0}=\begin{bmatrix} 2\\ 1 \end{bmatrix}

we will find dxdt\frac{dx}{dt}, AxAx, and verify that they are equal. After that we will compute

x(0)=[x1(0)x2(0)]x(0)=\begin{bmatrix} x_{1}(0)\\ x_{2}(0) \end{bmatrix}

to verify that the initial condition is satisfied as well.

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Biology 1st Edition by Kenneth R. Miller, Levine

Biology

1st EditionISBN: 9780132013499 (4 more)Kenneth R. Miller, Levine
2,470 solutions
Biology 1st Edition by Kenneth R. Miller, Levine

Biology

1st EditionISBN: 9780133669510 (5 more)Kenneth R. Miller, Levine
2,590 solutions
Miller and Levine Biology 1st Edition by Joseph S. Levine, Kenneth R. Miller

Miller and Levine Biology

1st EditionISBN: 9780328925124 (1 more)Joseph S. Levine, Kenneth R. Miller
1,773 solutions
Biocalculus: Calculus for the Life Sciences 1st Edition by Day, Stewart

Biocalculus: Calculus for the Life Sciences

1st EditionISBN: 9781133109631 (1 more)Day, Stewart
5,056 solutions

More related questions

1/4

1/7