Linear System of ODEs Solver (2x2)

Published on: July 18, 2026
Final Answer: Free • Full Steps: Pro

This Linear System ODE Solver solves a 2×2 first-order system x' = A x using the eigenvalue method. It finds the eigenvalues and eigenvectors of the coefficient matrix A and combines them into the general solution.

Step-by-step method

  1. Write the system in matrix form x' = A x.
  2. Use the eigenvalue method: x(t) is a combination of e^(λ t) v.
  3. Find the eigenvalues of A.
  4. Find an eigenvector for each eigenvalue.
  5. Combine them into the general solution.

Formula:

\(\vec{x}(t) = C_{1} e^{\lambda_{1} t} \vec{v}_{1} + C_{2} e^{\lambda_{2} t} \vec{v}_{2} + \cdots\)

Example 1:

\\( ec{x}\,' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} ec{x}\\)

Step 1 - Write the system in matrix form x' = A x.

In this problem: The system in matrix form is \( ec{x}\,' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} ec{x}\), with \(A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\).

\( ec{x}\,' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} ec{x}\)

Step 2 - Use the eigenvalue method: x(t) is a combination of e^(λ t) v.

In this problem: Each eigenvalue and eigenvector pair contributes one term to the solution.

\(\vec{x}(t) = C_{1} e^{\lambda_{1} t} \vec{v}_{1} + C_{2} e^{\lambda_{2} t} \vec{v}_{2} + \cdots\)

Step 3 - Find the eigenvalues of A.

In this problem: Solving det(A - \(\lambda\) I) = 0 gives \(\lambda = -1,\; \lambda = 4\).

\(\lambda = -1,\; \lambda = 4\)

Step 4a - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = -1\), an eigenvector is \(\left[\begin{matrix}-1\\1\end{matrix}\right]\).

\(\lambda = -1:\; ec{v} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\)

Step 4b - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = 4\), an eigenvector is \(\left[\begin{matrix}\frac{2}{3}\\1\end{matrix}\right]\).

\(\lambda = 4:\; ec{v} = \begin{pmatrix} \frac{2}{3} \\ 1 \end{pmatrix}\)

Step 5 - Combine them into the general solution.

In this problem: Combine the eigenvalue terms into the general solution with constants C1, C2, and so on.

\(egin{gathered} x_{1}(t) = \left(- C_{1} + \frac{2 C_{2} e^{5 t}}{3}\right) e^{- t} \\ x_{2}(t) = \left(C_{1} + C_{2} e^{5 t}\right) e^{- t} \end{gathered}\)

Final answer:

\\(egin{gathered} x_{1}(t) = \left(- C_{1} + \frac{2 C_{2} e^{5 t}}{3}\right) e^{- t} \\ x_{2}(t) = \left(C_{1} + C_{2} e^{5 t}\right) e^{- t} \end{gathered}\\)

Example 2:

\\( ec{x}\,' = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} ec{x}\\)

Step 1 - Write the system in matrix form x' = A x.

In this problem: The system in matrix form is \( ec{x}\,' = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} ec{x}\), with \(A = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}\).

\( ec{x}\,' = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} ec{x}\)

Step 2 - Use the eigenvalue method: x(t) is a combination of e^(λ t) v.

In this problem: Each eigenvalue and eigenvector pair contributes one term to the solution.

\(\vec{x}(t) = C_{1} e^{\lambda_{1} t} \vec{v}_{1} + C_{2} e^{\lambda_{2} t} \vec{v}_{2} + \cdots\)

Step 3 - Find the eigenvalues of A.

In this problem: Solving det(A - \(\lambda\) I) = 0 gives \(\lambda = 2 - i,\; \lambda = 2 + i\).

\(\lambda = 2 - i,\; \lambda = 2 + i\)

Step 4a - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = 2 - i\), an eigenvector is \(\left[\begin{matrix}- i\\1\end{matrix}\right]\).

\(\lambda = 2 - i:\; ec{v} = \begin{pmatrix} - i \\ 1 \end{pmatrix}\)

Step 4b - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = 2 + i\), an eigenvector is \(\left[\begin{matrix}i\\1\end{matrix}\right]\).

\(\lambda = 2 + i:\; ec{v} = \begin{pmatrix} i \\ 1 \end{pmatrix}\)

Step 5 - Combine them into the general solution.

In this problem: Combine the eigenvalue terms into the general solution with constants C1, C2, and so on.

\(egin{gathered} x_{1}(t) = - \left(C_{1} \sin{\left(t \right)} + C_{2} \cos{\left(t \right)}\right) e^{2 t} \\ x_{2}(t) = \left(C_{1} \cos{\left(t \right)} - C_{2} \sin{\left(t \right)}\right) e^{2 t} \end{gathered}\)

Final answer:

\\(egin{gathered} x_{1}(t) = - \left(C_{1} \sin{\left(t \right)} + C_{2} \cos{\left(t \right)}\right) e^{2 t} \\ x_{2}(t) = \left(C_{1} \cos{\left(t \right)} - C_{2} \sin{\left(t \right)}\right) e^{2 t} \end{gathered}\\)
See Example 1 Hide Example 1

Problem

\( ec{x}\,' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} ec{x}\)

Approach

Step-by-step method

  1. Write the system in matrix form x' = A x.
  2. Use the eigenvalue method: x(t) is a combination of e^(λ t) v.
  3. Find the eigenvalues of A.
  4. Find an eigenvector for each eigenvalue.
  5. Combine them into the general solution.

Step 1

Step 1 - Write the system in matrix form x' = A x.

In this problem: The system in matrix form is \( ec{x}\,' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} ec{x}\), with \(A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\).

\( ec{x}\,' = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} ec{x}\)

Step 2

Step 2 - Use the eigenvalue method: x(t) is a combination of e^(λ t) v.

In this problem: Each eigenvalue and eigenvector pair contributes one term to the solution.

\(\vec{x}(t) = C_{1} e^{\lambda_{1} t} \vec{v}_{1} + C_{2} e^{\lambda_{2} t} \vec{v}_{2} + \cdots\)

Step 3

Step 3 - Find the eigenvalues of A.

In this problem: Solving det(A - \(\lambda\) I) = 0 gives \(\lambda = -1,\; \lambda = 4\).

\(\lambda = -1,\; \lambda = 4\)

Step 4a

Step 4a - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = -1\), an eigenvector is \(\left[\begin{matrix}-1\\1\end{matrix}\right]\).

\(\lambda = -1:\; ec{v} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\)

Step 4b

Step 4b - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = 4\), an eigenvector is \(\left[\begin{matrix}\frac{2}{3}\\1\end{matrix}\right]\).

\(\lambda = 4:\; ec{v} = \begin{pmatrix} \frac{2}{3} \\ 1 \end{pmatrix}\)

Step 5

Step 5 - Combine them into the general solution.

In this problem: Combine the eigenvalue terms into the general solution with constants C1, C2, and so on.

\(egin{gathered} x_{1}(t) = \left(- C_{1} + \frac{2 C_{2} e^{5 t}}{3}\right) e^{- t} \\ x_{2}(t) = \left(C_{1} + C_{2} e^{5 t}\right) e^{- t} \end{gathered}\)

Final Answer

\(egin{gathered} x_{1}(t) = \left(- C_{1} + \frac{2 C_{2} e^{5 t}}{3}\right) e^{- t} \\ x_{2}(t) = \left(C_{1} + C_{2} e^{5 t}\right) e^{- t} \end{gathered}\)
See Example 2 Hide Example 2

Problem

\( ec{x}\,' = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} ec{x}\)

Approach

Step-by-step method

  1. Write the system in matrix form x' = A x.
  2. Use the eigenvalue method: x(t) is a combination of e^(λ t) v.
  3. Find the eigenvalues of A.
  4. Find an eigenvector for each eigenvalue.
  5. Combine them into the general solution.

Step 1

Step 1 - Write the system in matrix form x' = A x.

In this problem: The system in matrix form is \( ec{x}\,' = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} ec{x}\), with \(A = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}\).

\( ec{x}\,' = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix} ec{x}\)

Step 2

Step 2 - Use the eigenvalue method: x(t) is a combination of e^(λ t) v.

In this problem: Each eigenvalue and eigenvector pair contributes one term to the solution.

\(\vec{x}(t) = C_{1} e^{\lambda_{1} t} \vec{v}_{1} + C_{2} e^{\lambda_{2} t} \vec{v}_{2} + \cdots\)

Step 3

Step 3 - Find the eigenvalues of A.

In this problem: Solving det(A - \(\lambda\) I) = 0 gives \(\lambda = 2 - i,\; \lambda = 2 + i\).

\(\lambda = 2 - i,\; \lambda = 2 + i\)

Step 4a

Step 4a - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = 2 - i\), an eigenvector is \(\left[\begin{matrix}- i\\1\end{matrix}\right]\).

\(\lambda = 2 - i:\; ec{v} = \begin{pmatrix} - i \\ 1 \end{pmatrix}\)

Step 4b

Step 4b - Find an eigenvector for each eigenvalue.

In this problem: For \(\lambda = 2 + i\), an eigenvector is \(\left[\begin{matrix}i\\1\end{matrix}\right]\).

\(\lambda = 2 + i:\; ec{v} = \begin{pmatrix} i \\ 1 \end{pmatrix}\)

Step 5

Step 5 - Combine them into the general solution.

In this problem: Combine the eigenvalue terms into the general solution with constants C1, C2, and so on.

\(egin{gathered} x_{1}(t) = - \left(C_{1} \sin{\left(t \right)} + C_{2} \cos{\left(t \right)}\right) e^{2 t} \\ x_{2}(t) = \left(C_{1} \cos{\left(t \right)} - C_{2} \sin{\left(t \right)}\right) e^{2 t} \end{gathered}\)

Final Answer

\(egin{gathered} x_{1}(t) = - \left(C_{1} \sin{\left(t \right)} + C_{2} \cos{\left(t \right)}\right) e^{2 t} \\ x_{2}(t) = \left(C_{1} \cos{\left(t \right)} - C_{2} \sin{\left(t \right)}\right) e^{2 t} \end{gathered}\)
Coefficient matrix A