Solving Linear Differential Equations

Solving Linear Differential Equations:

  1. Homogeneous Linear Differential Equations:

    • Characteristic Equation Method:
      • Assume the solution of the form y=erx.
      • Substitute this solution into the homogeneous equation to form a characteristic equation using derivatives.
      • Solve the characteristic equation to find roots r1,r2,,rn.
      • The general solution is y=C1er1x+C2er2x++Cnernx, where C1,C2,,Cnare constants.
  2. Non-Homogeneous Linear Differential Equations:

    • Method of Undetermined Coefficients:

      • For the non-homogeneous equation an(x)dnydxn+an1(x)dn1ydxn1++a1(x)dydx+a0(x)y=g(x):
      • Solve the associated homogeneous equation to find the complementary function (CF).
      • Guess a form for the particular solution (PS) based on g(x) and its derivatives.
      • Substitute the guessed form into the non-homogeneous equation, solve for undetermined coefficients, and add the CF to get the general solution.
    • Variation of Parameters:

      • Applicable when the non-homogeneous term g(x) is expressed as a function.
      • Assume the form of a particular solution yp(x) with undetermined coefficients involving the complementary function.
      • Determine the variation parameters involving integrals of functions derived from the homogeneous solution.
      • Combine the CF with the PS obtained through variation of parameters to form the general solution.
  • Example: Solving a Linear Differential Equation by Undetermined Coefficients:

    Consider the equation y4y=3x2+2ex.

    a. Solve the Associated Homogeneous Equation:

    • The associated homogeneous equation is y4y=0, leading to the CF: yCF=C1e2x+C2e2x.

    b. Guess a Form for the Particular Solution (PS):

    • Based on 3x2 and 2ex, assume yp(x)=Ax2+Bex, where A and B are undetermined coefficients.

    c. Substitute the Guessed Form into the Non-Homogeneous Equation:

    • Find y and y for yp(x) and substitute into the non-homogeneous equation.
    • Equate coefficients of similar terms on both sides to solve for A and B.

    d. Combine the CF with the PS to Form the General Solution:

    • The general solution is y=C1e2x+C2e2x+Ax2+Bex, where A and B are the determined coefficients.
  • Example of Solving a Linear Differential Equation:

    Consider the differential equation: y4y+4y=3e2x.

    a. Homogeneous Solution (CF):

    • The associated homogeneous equation is y4y+4y=0.
    • Characteristic equation: r24r+4=0 gives a repeated root r=2.
    • The CF is yCF=(C1+C2x)e2x.

    b. Particular Solution (PS) by Method of Undetermined Coefficients:

    • Assume a particular solution of the form yPS=Ae2x for the non-homogeneous equation.
    • Substitute into the equation and solve for A: 3e2x=3Ae2x. Therefore, A=1.
    • The PS is yPS=e2x.

    c. General Solution:

    • The general solution is the sum of CF and PS: y=yCF+yPS=(C1+C2x)e2x+e2x.