Method of undetermined coefficients

Method of undetermined coefficients

Lessons

A non-homogeneous constant coefficient second order linear differential equation is of the form:
Ay+By+Cy=d(x)Ay''+By'+Cy=d(x)

The complementary solution to Ay+By+Cy=0Ay''+By'+Cy=0 is yc(x)=c1f(x)+c2g(x)y_c (x)=c_1 f(x)+c_2 g(x)

And the particular solution to Ay+By+Cy=d(x)Ay''+By'+Cy=d(x) is yp(x)y_p (x)

So the full general solution to
Ay+By+Cy=d(x)Ay''+By'+Cy=d(x)

Will be y(x)=yc+ypy(x)=y_c+y_p

If yp1(x)y_p1 (x) is a particular solution for
Ay+By+Cy=d1(x)Ay''+By'+Cy=d_1 (x)

And yp2(x)y_p2 (x) is a particular solution for
Ay+By+Cy=d2(x)Ay''+By'+Cy=d_2 (x)

Then yp1(x)+yp2(x)y_p1 (x)+y_p2 (x) is a solution to
Ay+By+Cy=d1(x)+d2(x)Ay''+By'+Cy=d_1 (x)+d_2 (x)


table of particular solutions of different d(x)
  • Introduction
    a)
    What is the Method of Undetermined Coefficients?

    b)
    How to find sums of particular solutions


  • 1.
    Using the Method of Undetermined Coefficients
    Find the solution to the following differential equation:

    y6y+5y=2e3xy''-6y'+5y=2e^{3x}

    With initial values y(0)=12y(0)=-\frac{1}{2}, and y(0)=12y' (0)=-\frac{1}{2}

  • 2.
    Find the solution to the following differential equation:

    y+y6y=4sin(2x)y''+y'-6y=4\sin(2x)

    With initial values y(0)=1213y(0)=\frac{12}{13}, and y(0)=1013y' (0)=-\frac{10}{13}

  • 3.
    Find the solution to the following differential equation:

    y+y6y=12x3+3x2+196y''+y'-6y=-12x^3+3x^2+\frac{19}{6}

    With initial values y(0)=3y(0)=3 and y(0)=1y' (0)=-1

  • 4.
    Find the general solution to the following differential equation:
    y+2y+y=xe3x+5cos(3x)y''+2y'+y=xe^{3x}+5\cos(3x)