Calculator
Example Data Table
This sample shows a damped, forced system. You can use these values directly in the form to test the solver and compare exact and RK4 outputs.
| a11 | a12 | a21 | a22 | c1 | c2 | x0 | y0 | t0 | t_eval | t_end | points | substeps |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 1 | -2 | -3 | 0 | 4 | 1 | 0 | 0 | 4 | 10 | 21 | 12 |
| 2 | -1 | 3 | -4 | 1 | -2 | 0.5 | 1.5 | 0 | 2 | 6 | 25 | 15 |
Formula Used
For a two-variable linear system, the calculator models:
where
When the coefficient matrix is invertible, the exact solution is computed with the matrix exponential:
Here, Xp = -A⁻¹b is the equilibrium or particular solution, and X0 is the initial vector at time t0.
The page also computes a Runge–Kutta fourth-order approximation, so you can compare a numerical method against the exact matrix-based result.
How to Use This Calculator
- Enter the four matrix coefficients for the coupled system.
- Enter the constant forcing terms c1 and c2.
- Provide initial values x(t0), y(t0), and the initial time t0.
- Choose the target time for the main result.
- Set the plot end time, sample count, and RK4 substeps.
- Click Solve System to place the result above the form.
- Review stability, eigenvalues, equilibrium, table values, and the graph.
- Use the export buttons to save the current results as CSV or PDF.
Frequently Asked Questions
1) What kind of systems does this solver handle?
It solves two first-order linear simultaneous differential equations with constant coefficients and constant forcing terms. It is designed for coupled x(t) and y(t) systems.
2) Does it return exact values or numerical values?
It returns exact matrix-exponential values when the coefficient matrix is invertible. It also computes RK4 numerical values for comparison and for singular-matrix fallback cases.
3) Why are eigenvalues shown?
Eigenvalues describe growth, decay, oscillation, and stability. They help classify the system as a node, spiral, saddle, or neutral oscillatory pattern.
4) What does the equilibrium point mean?
The equilibrium is the steady state where both derivatives become zero. In a stable system, nearby solutions often move toward this point over time.
5) Why compare exact and RK4 values?
The comparison shows numerical accuracy. Small differences mean the selected sampling and substeps are adequate for the interval you chose.
6) What should I do if the errors seem large?
Increase the RK4 substeps or use more sample points. That reduces step size and usually improves the numerical approximation.
7) Can this solver work backward in time?
Yes. You can use a target time or plot end time smaller than the initial time. The integration step simply becomes negative.
8) Is this page suitable for nonlinear systems?
No. This version is built for linear two-variable systems with constant coefficients. Nonlinear systems need a different solver and different stability rules.