Question

The solution of the system of ODEs: \(\frac{dx}{dt}=x+2y\), \(\frac{dy}{dt}=3x+2y\) with initial conditions x(0)=6 and y(0)=4 is ____.

• A. \(x(t)=4e^{4t}+2e^t,\ y(t)=6e^{4t}-2e^t\)
• B. \(x(t)=4e^{4t}+2e^{-t},\ y(t)=6e^{4t}-2e^{-t}\)
• C. \(x(t)=6e^{4t}-2e^{-t},\ y(t)=2e^{4t}+2e^{-t}\)
• D. \(x(t)=2e^{4t}+4e^{-t},\ y(t)=2e^{4t}+2e^{-t}\)

Hint:

Find the eigenvalues of the coefficient matrix (4 and -1), form the general solution, then fit the constants to x(0)=6, y(0)=4.

Answer 1

The Correct Option is B

Step 1: Write the system in matrix form \(\mathbf{X}' = A\mathbf{X}\) with

\[A = \begin{pmatrix}1 & 2\\ 3 & 2\end{pmatrix}, \qquad \mathbf{X} = \begin{pmatrix}x\\ y\end{pmatrix}\]

Step 2: Find the eigenvalues from \(\det(A-\lambda I)=0\):

\[(1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = (\lambda-4)(\lambda+1) = 0\]

so \(\lambda_1 = 4\) and \(\lambda_2 = -1\).

Step 3: For \(\lambda_1 = 4\), solve \((A-4I)\mathbf{v}=0\):

\[\begin{pmatrix}-3 & 2\\ 3 & -2\end{pmatrix}\mathbf{v}=0 \implies -3v_1+2v_2=0 \implies \mathbf{v}_1=\begin{pmatrix}2\\3\end{pmatrix}\]

Step 4: For \(\lambda_2 = -1\), solve \((A+I)\mathbf{v}=0\):

\[\begin{pmatrix}2 & 2\\ 3 & 3\end{pmatrix}\mathbf{v}=0 \implies v_1+v_2=0 \implies \mathbf{v}_2=\begin{pmatrix}1\\-1\end{pmatrix}\]

Step 5: The general solution is

\[\begin{pmatrix}x(t)\\y(t)\end{pmatrix} = c_1\begin{pmatrix}2\\3\end{pmatrix}e^{4t} + c_2\begin{pmatrix}1\\-1\end{pmatrix}e^{-t}\]

so \(x(t) = 2c_1e^{4t}+c_2e^{-t}\) and \(y(t) = 3c_1e^{4t}-c_2e^{-t}\).

Step 6: Apply the initial conditions \(x(0)=6\), \(y(0)=4\):

\[2c_1+c_2=6, \qquad 3c_1-c_2=4\]

Adding the two equations gives \(5c_1=10\), so \(c_1=2\), and then \(c_2=6-2c_1=2\).

Step 7: Substitute \(c_1=2\), \(c_2=2\) back:

\[x(t) = 4e^{4t}+2e^{-t}, \qquad y(t) = 6e^{4t}-2e^{-t}\] \[\boxed{x(t)=4e^{4t}+2e^{-t},\ y(t)=6e^{4t}-2e^{-t}}\]