Math Problem Statement

The problem involves solving a system of differential equations by applying Euler's formula to the complex exponential solution given for \(\vec{Y}(t)\).

Solution

The problem involves solving a system of differential equations by applying Euler's formula to the complex exponential solution given for Y(t)\vec{Y}(t).

Problem Overview:

We are given that: Y(t)=e(2+6i)t[37i]\vec{Y}(t) = e^{(2 + 6i)t} \begin{bmatrix} -3 \\ 7i \end{bmatrix} is a solution to the system: dYdt=AY\frac{d\vec{Y}}{dt} = A \vec{Y} The question asks which expression results from applying Euler's formula.

Euler's Formula:

Euler's formula states that for any real number θ\theta: eiθ=cos(θ)+isin(θ)e^{i\theta} = \cos(\theta) + i \sin(\theta)

Solution Breakdown:

The solution Y(t)\vec{Y}(t) can be broken down using Euler's formula: e(2+6i)t=e2te6ite^{(2 + 6i)t} = e^{2t} \cdot e^{6it} Using Euler’s formula for e6ite^{6it}, we have: e6it=cos(6t)+isin(6t)e^{6it} = \cos(6t) + i\sin(6t) Thus, e(2+6i)t=e2t(cos(6t)+isin(6t))e^{(2 + 6i)t} = e^{2t} (\cos(6t) + i \sin(6t))

Now, we apply this to the vector [37i]\begin{bmatrix} -3 \\ 7i \end{bmatrix}: Y(t)=e2t(cos(6t)+isin(6t))[37i]\vec{Y}(t) = e^{2t} \left(\cos(6t) + i \sin(6t)\right) \begin{bmatrix} -3 \\ 7i \end{bmatrix}

The real and imaginary components of the vector will be handled separately. Breaking it down:

Real part:

Multiplying the real part cos(6t)\cos(6t) by the vector: cos(6t)[37i]=[3cos(6t)7icos(6t)]=[3cos(6t)7sin(6t)]\cos(6t) \begin{bmatrix} -3 \\ 7i \end{bmatrix} = \begin{bmatrix} -3\cos(6t) \\ 7i\cos(6t) \end{bmatrix} = \begin{bmatrix} -3\cos(6t) \\ -7\sin(6t) \end{bmatrix} (since i2=1i^2 = -1)

Imaginary part:

Multiplying the imaginary part isin(6t)i\sin(6t) by the vector: isin(6t)[37i]=[3isin(6t)7(1)sin(6t)]=[3isin(6t)7cos(6t)]i\sin(6t) \begin{bmatrix} -3 \\ 7i \end{bmatrix} = \begin{bmatrix} -3i\sin(6t) \\ 7(-1)\sin(6t) \end{bmatrix} = \begin{bmatrix} -3i\sin(6t) \\ -7\cos(6t) \end{bmatrix}

Thus, Y(t)\vec{Y}(t) becomes: Y(t)=e2t[[3cos(6t)7sin(6t)]+i[3sin(6t)7cos(6t)]]\vec{Y}(t) = e^{2t} \left[ \begin{bmatrix} -3\cos(6t) \\ -7\sin(6t) \end{bmatrix} + i \begin{bmatrix} -3\sin(6t) \\ -7\cos(6t) \end{bmatrix} \right]

Conclusion:

This matches the third option (from the top) on the list in the image: Y=e2t[3cos(6t)7sin(6t)]+ie2t[3sin(6t)7cos(6t)]\vec{Y} = e^{2t} \begin{bmatrix} -3\cos(6t) \\ 7\sin(6t) \end{bmatrix} + i e^{2t} \begin{bmatrix} -3\sin(6t) \\ -7\cos(6t) \end{bmatrix}

Would you like any further explanation or details?


Here are five related questions for further exploration:

  1. How does Euler’s formula relate to solving complex differential equations?
  2. Can you explain the process of separating real and imaginary parts in complex functions?
  3. How would the solution change if the coefficient of ii in the exponent were different?
  4. What role does the matrix AA play in the system dYdt=AY\frac{d\vec{Y}}{dt} = A\vec{Y}?
  5. How can we solve this problem using matrix exponentiation?

Tip: Always break down complex exponential expressions into real and imaginary parts to simplify the analysis and identify real-world implications.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Euler's Formula
Complex Numbers
Linear Algebra

Formulas

e^{i\theta} = \cos(\theta) + i \sin(\theta)
e^{(a + bi)t} = e^{at} \cdot e^{i bt}

Theorems

Euler's Formula

Suitable Grade Level

Undergraduate