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Exercise 4: Find the solution to the initial value problem \[ \vec{X}^{\prime}=\left(\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right) \vec{X}+\binom{1}{-e^{2 t}}, \quad \vec{X}(0)=\binom{3}{1} \]

          Exercise 4: Find the solution to the initial value problem
\[
\vec{X}^{\prime}=\left(\begin{array}{rr}
3 & -4 \\
1 & -1
\end{array}\right) \vec{X}+\binom{1}{-e^{2 t}}, \quad \vec{X}(0)=\binom{3}{1}
\]

Exercise 4: Find the solution to the initial value problem

X⃗^'=(
3 -4

1 -1
) X⃗+1-e^2 t, X⃗(0)=31

Added by Sergio W.

Calculus and Its Applications

Larry J. Goldstein, David C. Lay, David… 14th Edition

Chapter 10

Instant Answer

Solved by Expert Linda Hand

Step 1

\[ \vec{X}^{\prime} = \left(\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right) \vec{X} + \binom{1}{-e^{2t}}, \quad \vec{X}(0) = \binom{3}{1} \] Show more…

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Exercise 4: Find the solution to the initial value problem \[ \vec{X}^{\prime}=\left(\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right) \vec{X}+\binom{1}{-e^{2 t}}, \quad \vec{X}(0)=\binom{3}{1} \]