Question:
The solution \(y(x) = \binom{y_1(x)}{y_2(x)}\) of the initial value problem
\[
\frac{dy}{dx} = \begin{pmatrix} -3 & 4 \\ -2 & 3 \end{pmatrix} y, \qquad y(0) = \binom{1}{2}
\]
is equal to
The solution \(y(x) = \binom{y_1(x)}{y_2(x)}\) of the initial value problem
\[
\frac{dy}{dx} = \begin{pmatrix} -3 & 4 \\ -2 & 3 \end{pmatrix} y, \qquad y(0) = \binom{1}{2}
\]
is equal to
Show Hint
Find the eigenvalues and eigenvectors of the coefficient matrix, then fix the constants using the given initial vector \(y(0)=\binom{1}{2}\).
Updated On: Jul 21, 2026
- \(3\dbinom{1}{1}e^{x} + \dbinom{2}{1}e^{-x}\)
- \(-3\dbinom{1}{1}e^{x} + \dbinom{2}{1}e^{-x}\)
- \(3\dbinom{1}{1}e^{x} + 2\dbinom{2}{1}e^{-x}\)
- \(3\dbinom{1}{1}e^{x} - \dbinom{2}{1}e^{-x}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Find the eigenvalues of the coefficient matrix.
The system is \(y' = Ay\) with \(A = \begin{pmatrix} -3 & 4 \\ -2 & 3 \end{pmatrix}\). The trace of \(A\) is \(-3+3=0\) and the determinant is \((-3)(3)-(4)(-2)=-9+8=-1\).
The characteristic equation is \(\lambda^2 - (\text{trace})\lambda + \det = 0\), so \(\lambda^2 - 1 = 0\), which gives \(\lambda = 1\) and \(\lambda = -1\).
Step 2: Find the eigenvector for each eigenvalue.
For \(\lambda = 1\): solve \((A-I)v=0\), that is \(\begin{pmatrix} -4 & 4 \\ -2 & 2 \end{pmatrix}v=0\). This gives \(v_1 = v_2\), so take \(v=\binom{1}{1}\).
For \(\lambda = -1\): solve \((A+I)v=0\), that is \(\begin{pmatrix} -2 & 4 \\ -2 & 4 \end{pmatrix}v=0\). This gives \(v_1 = 2v_2\), so take \(v=\binom{2}{1}\).
Step 3: Write the general solution and use the initial condition.
Since the eigenvalues are real and distinct, the general solution is \(y(x)=c_1\binom{1}{1}e^{x}+c_2\binom{2}{1}e^{-x}\).
At \(x=0\): \(c_1\binom{1}{1}+c_2\binom{2}{1}=\binom{1}{2}\), which gives \(c_1+2c_2=1\) and \(c_1+c_2=2\).
Subtracting the second equation from the first gives \(c_2=-1\), and then \(c_1=2-c_2=3\).
Step 4: Check the wrong options.
Option (A) uses \(c_1=3,c_2=1\), which gives \(y(0)=\binom{5}{4}\), not \(\binom{1}{2}\), so it is wrong.
Option (B) uses \(c_1=-3,c_2=1\), which gives \(y(0)=\binom{-1}{-2}\), the negative of the required vector, so it is wrong.
Option (C) uses \(c_1=3,c_2=2\), which gives \(y(0)=\binom{7}{5}\), not \(\binom{1}{2}\), so it is also wrong.
Final Answer:
With \(c_1=3\) and \(c_2=-1\), the solution is \(y(x)=3\binom{1}{1}e^{x}-\binom{2}{1}e^{-x}\). \[ \boxed{y(x)=3\binom{1}{1}e^{x}-\binom{2}{1}e^{-x}} \]
The system is \(y' = Ay\) with \(A = \begin{pmatrix} -3 & 4 \\ -2 & 3 \end{pmatrix}\). The trace of \(A\) is \(-3+3=0\) and the determinant is \((-3)(3)-(4)(-2)=-9+8=-1\).
The characteristic equation is \(\lambda^2 - (\text{trace})\lambda + \det = 0\), so \(\lambda^2 - 1 = 0\), which gives \(\lambda = 1\) and \(\lambda = -1\).
Step 2: Find the eigenvector for each eigenvalue.
For \(\lambda = 1\): solve \((A-I)v=0\), that is \(\begin{pmatrix} -4 & 4 \\ -2 & 2 \end{pmatrix}v=0\). This gives \(v_1 = v_2\), so take \(v=\binom{1}{1}\).
For \(\lambda = -1\): solve \((A+I)v=0\), that is \(\begin{pmatrix} -2 & 4 \\ -2 & 4 \end{pmatrix}v=0\). This gives \(v_1 = 2v_2\), so take \(v=\binom{2}{1}\).
Step 3: Write the general solution and use the initial condition.
Since the eigenvalues are real and distinct, the general solution is \(y(x)=c_1\binom{1}{1}e^{x}+c_2\binom{2}{1}e^{-x}\).
At \(x=0\): \(c_1\binom{1}{1}+c_2\binom{2}{1}=\binom{1}{2}\), which gives \(c_1+2c_2=1\) and \(c_1+c_2=2\).
Subtracting the second equation from the first gives \(c_2=-1\), and then \(c_1=2-c_2=3\).
Step 4: Check the wrong options.
Option (A) uses \(c_1=3,c_2=1\), which gives \(y(0)=\binom{5}{4}\), not \(\binom{1}{2}\), so it is wrong.
Option (B) uses \(c_1=-3,c_2=1\), which gives \(y(0)=\binom{-1}{-2}\), the negative of the required vector, so it is wrong.
Option (C) uses \(c_1=3,c_2=2\), which gives \(y(0)=\binom{7}{5}\), not \(\binom{1}{2}\), so it is also wrong.
Final Answer:
With \(c_1=3\) and \(c_2=-1\), the solution is \(y(x)=3\binom{1}{1}e^{x}-\binom{2}{1}e^{-x}\). \[ \boxed{y(x)=3\binom{1}{1}e^{x}-\binom{2}{1}e^{-x}} \]
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