Method of Constant Coefficients

The method of constant Coefficients can be used to find the homogeneous solution to a linear differential equation with constant coefficients of any order.

The Essentials

A homogeneous linear differential equation with constant coefficients fits into the form:

\[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0 \]

The solutions will fit into this form: \( e^{rx} \) where r can be found from the corresponding equation:

\[ a_n r^{(n)} + a_{(n-1)} r^{(n-1)} + \cdots + a_1 r + a_0 = 0 \]

Which can be solved by factoring. After finding the different roots which make the above equation equal 0, there are four factorization cases that might apply to the solution:

Case 1: Distinct Real Roots

If the roots to the equation are different and real, the solution is simple:

\[ y(x) = C_1 e^{r_1 x} + \ldots + C_n e^{r_n x} \]

Case 2: Repeated Real Roots

If any of the roots are repeated, another term is added to the solution for each repeat with an additional coefficient of x. For example, if our roots are: \( r_1, \) \( \color{red}{r_2} \), \( \color{red}{r_2}\), \( \color{red}{r_2}\), \( r_3 \), then our solution is:

\[ C_1 e^{r_1 x} + C_2 \textcolor{red}{ e^{r_2 x} } + C_3 \textcolor{red}{ xe^{r_2 x} } + C_4 \textcolor{red}{ x^2 e^{r_2 x} } + C_5 e^{r_3 x} \]

Case 3: Complex Roots

If there are complex roots, they come in pairs. If two of our roots are: \( r = \lambda \pm \omega i \), then the two corresponding terms in the solution are:

\[ C_1 e^{\lambda x} \cos(\omega x) + C_2 e^{\lambda x} \sin(\omega x) \]

Case 4: Repeated Complex Roots

Just like in the case of repeated real roots, when complex roots are repeated, another two terms are added, a sin and a cos term, each multiplied by x for each repeat. If the roots r = λ±ωi are repeated three times, then these six terms are included in the solution:

\[ C_1 e^{\lambda x} \cos(\omega x) + C_2 e^{\lambda x} \sin(\omega x) + C_3 \textcolor{red}{x} e^{\lambda x} \cos(\omega x) + C_4 \textcolor{red}{x} e^{\lambda x} \sin(\omega x) + C_5 \textcolor{red}{x^2} e^{\lambda x} \cos(\omega x) + C_6 \textcolor{red}{x^2} e^{\lambda x} \sin(\omega x) \]

Note that every polynomial of order n can be factored, and will have n roots.

Example

Solve the equation:

\[ y^{(4)} - 2y^{(3)} + 2y'' - 2y' + y = 0 \]

To find the roots, we exchange the nth derivative of y with the nth power of r:

\[ r^4 - 2r^3 + 2r^2 - 2r + 1 = 0 \]

To factor, we can split the middle term:

\[ (r^4 - 2r^3 + r^2) + (r^2 - 2r + 1) = 0 \] \[ r^2(r^2 - 2r + 1) + (r^2 - 2r + 1) = 0 \] \[ (r^2 - 2r + 1)(r^2 + 1) = 0 \] \[ (r - 1)^2(r^2 + 1) = 0 \] \[ r = 1, 1, 0 \pm i \] \[ y(x) = C_1 e^{(1)x} + C_2 x e^{(1)x} + C_3 e^{0x} \sin((1)x) + C_4 e^{0x} \cos((1)x) \] \[ y(x) = C_1 e^x + C_2 x e^x + C_3 \sin(x) + C_4 \cos(x) \]

Practice

Evaluate these expressions:

  1. \( y''' - y'' + y' - y = 0 \)
  2. \( y^{(4)} + 8y'' + 16y = 0 \)
  3. \( y'' - 4y' + 5y = 0 \)

Solution:

  1. \( y(x) = C_1 e^x + C_2 \cos(x) + C_3 \sin(x) \)
  2. \( y(x) = C_1 \cos(2x) + C_2 \sin(2x) + C_3 x \cos(2x) + C_4 x \sin(2x) \)
  3. \( y(x) = C_1 e^{2x} \cos(x) + C_2 e^{2x} \sin(x) \)