The differential equation: y' (t) − y(t) = 0 can be written in a vector equation: ´y' (t) 1 'y' (t) [y(t) 9] Ly(t) O True O False
The differential equation: y' (t) − y(t) = 0 can be written in a vector equation: ´y' (t) 1 'y' (t) [y(t) 9] Ly(t) O True O False
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Differential Equations and Vector Equations**
The differential equation:
\[ y''(t) - y(t) = 0 \]
can be written in a vector equation form:
\[
\left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right]' = \left[\begin{array}{cc}
1 & 0 \\
1 & 0
\end{array}\right] \left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right].
\]
**True or False:**
- True
- False
In this question, you need to evaluate whether the given differential equation has been correctly transformed into the provided vector equation.
**Explanation of the Equation:**
The first-order system of differential equations is written as:
\[
\left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right]',
\]
which represents the derivative of the vector \( \left[\begin{array}{c} y'(t) \\ y(t) \end{array}\right] \).
The matrix multiplication on the right-hand side of the equation is:
\[
\left[\begin{array}{cc}
1 & 0 \\
1 & 0
\end{array}\right] \left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right].
\]
This represents a linear transformation of the vector \( \left[\begin{array}{c} y'(t) \\ y(t) \end{array}\right] \).
**Task:**
Evaluate the given vector equation by considering the structure of the differential equation and the corresponding vector form. Select "True" if the transformation is correct and "False" if it is incorrect.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F238584dd-2c47-447e-aaba-b9fce1ec1b5c%2F198e4317-4b32-46e7-bb64-e42dfbf14db1%2F22w8z3_processed.png&w=3840&q=75)
Transcribed Image Text:**Differential Equations and Vector Equations**
The differential equation:
\[ y''(t) - y(t) = 0 \]
can be written in a vector equation form:
\[
\left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right]' = \left[\begin{array}{cc}
1 & 0 \\
1 & 0
\end{array}\right] \left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right].
\]
**True or False:**
- True
- False
In this question, you need to evaluate whether the given differential equation has been correctly transformed into the provided vector equation.
**Explanation of the Equation:**
The first-order system of differential equations is written as:
\[
\left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right]',
\]
which represents the derivative of the vector \( \left[\begin{array}{c} y'(t) \\ y(t) \end{array}\right] \).
The matrix multiplication on the right-hand side of the equation is:
\[
\left[\begin{array}{cc}
1 & 0 \\
1 & 0
\end{array}\right] \left[\begin{array}{c}
y'(t) \\
y(t)
\end{array}\right].
\]
This represents a linear transformation of the vector \( \left[\begin{array}{c} y'(t) \\ y(t) \end{array}\right] \).
**Task:**
Evaluate the given vector equation by considering the structure of the differential equation and the corresponding vector form. Select "True" if the transformation is correct and "False" if it is incorrect.
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