How to Solve Simultaneous Differential Equations

Simultaneous differential equations are solved using a coefficient matrix to create an eigenvalue problem. Once the eigenvalues are calculated, they are reintroduced into the simultaneous equations to determine the general solution. A firm knowledge of integral calculus is required because you must first "guess" the form of the solution based solely on the construction of the equations in the problem. For instance, you must be able to see the equation y''+ay'+by = 0 and know that the solution takes the form of y=e^(lambda*t).

Things You'll Need

  • Pen/pencil Paper
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Instructions

    • 1

      Write the equations in standard form and create the coefficient matrix. The picture shows the coefficient matrix for the following example.
      y1' = 2*y1 - 4*y2
      y2' = 1*y1 - 3*y2

    • 2

      Subtract the eigenvalue lambda multiplied by the identity matrix from the coefficient matrix.

    • 3

      Calculate the eigenvalue equation for the determinant of the newly formed matrix and set it equal to zero.
      (2-lambda)(-3-lambda) - (-4*1)
      -6 + lambda + lambda^2 + 4
      lambda^2 + lambda - 2 = 0

    • 4

      Solve the equation to determine the eigenvalues.
      lambda^2 + lambda - 2 = 0
      (lambda - 1)(lambda + 2) = 0
      lambda1 = 1; lambda2 = -2

    • 5

      Use the eigenvalues and the eigenvalue matrix to determine the eigenvectors.
      (2 - lamda)*x1 - 4*x2 = 0
      x1 = 4x2; eigenvector is [4, 1] for lambda = 1
      4x1 = 4x2; eigenvector is [1, 1] for lambda = -2

    • 6

      Write the general solution using the eigenvalues and eigenvectors. For this example, the solution is in the form of y = e^(lambda*t) and since this is the general solution, arbitrary constants which are a result of integration are introduced.
      y1 = 4*c1*e^t + c2*e^(-2t)
      y2 = c1*e^t + c2*e^(-2t)

    • 7

      Use the initial or boundary conditions to obtain values for the arbitrary constants. This example is an initial condition problem. The initial conditions are y1(0) = 3 and y2(0) = 0.
      y1 = 4*c1*e^0 + c2*e^(-2*0) = 4*c1 + c2 = 3
      y2 = c1*e^0 + c2*e^(-2*0) = c1 + c2 = 0
      4*c1 = 3 - c2
      c1 = -c2
      4*(-c2) = 3 - c2; c2 = -1
      c1 = -(-1); c1 = 1

    • 8

      Write the particular solution by substituting the values for the constants back into the general solution:
      y1 = 4e^t - e^(-2t)
      y2 = e^t - e^(-2t)

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