Cauchy-Euler Method

The homogeneous solution to differential equations of any order with variable coefficients can be found if they are a Cauchy-Euler equation.

The Essentials

This is Cauchy-Euler form:

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

Solutions to this type of equation look like this: \( y(x) = x^r \) . This solution is differentiated and substituted into the equation to find what the values for r are:

\[ y = x^r \] \[ y' = r x^{r-1} \] \[ y'' = r(r - 1)x^{r-2} \] \[ y''' = r(r - 1)(r - 2)x^{r-3} \]

And so on. After that, factor out an xr and solve for the roots. Just like with the method of constant coefficients, there are four cases with roots:

Case 1: Distinct Real Roots

For different roots, each root is the exponent. If a second order differential equation is considered, and the two roots are \( r_1 \), \( r_2 \), then the solution is:

\[ y(x) = C_1 x^{r_1} + C_2 x^{r_2} \]

Case 2: Repeated Real Roots

For each root that is repeated, a natural log is multiplied to the term for each repeat. If the roots are \( r_1 \), \( r_2 \), \( r_2 \), \( r_2 \), \( r_3 \), then the solution would be:

\[ y(x) = C_1 x^{r_1} + C_2 x^{r_2} + C_3 x^{r_2} \ln(x) + C_4 x^{r_2} \ln^2(x) + C_5 x^{r_3} \]

Case 3: Complex Roots

Complex roots come in pairs: \( \lambda \pm \omega i \). For each pair, these two terms are added to the solution:

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

Case 4: Repeated Complex Roots

If there is a repeated complex pair then for each repeat, add a term for each repeat multiplied by the natural log for each repeat already written. If there are four roots that are: \( \lambda \pm \omega i, \lambda \pm \omega i \)

\[ y(x) = C_1 x^{\lambda} \cos(\omega \ln(x)) + C_2 x^{\lambda} \sin(\omega \ln(x)) + C_3 x^{\lambda} \ln(x) \cos(\omega \ln(x)) + C_4 x^{\lambda} \ln(x) \sin(\omega \ln(x)) \]

Example

Solve the equation:

\[ x^4 y^{(4)} + 6x^3 y''' + 9x^2 y'' + 3xy' + y = 0 \]

First, we differentiate the form for our solution \( y = x^r \):

\[ y = x^r \] \[ y' = rx^{r-1} \] \[ y'' = r(r - 1)x^{r-2} = (r^2 - r)x^{r-2} \] \[ y''' = r(r - 1)(r - 2)x^{r-3} = (r^3 - 3r^2 + 2r)x^{r-3} \] \[ y^{(4)} = r(r - 1)(r - 2)(r - 3)x^{r-4} = (r^4 - 6r^3 + 11r^2 - 6r)x^{r-4} \]

Plugging those into our original equation we get:

\[ x^4 y^{(4)} + 6x^3 y''' + 9x^2 y'' + 3xy' + y = 0 \] \[ x^4 x^{r-4}(r^4 - 6r^3 + 11r^2 - 6r) + 6x^3 x^{r-3}(r^3 - 3r^2 + 2r) + 9x^2 x^{r-2}(r^2 - r) + 3x x^{r-1} r + x^r = 0 \] \[ x^r(r^4 - 6r^3 + 11r^2 - 6r) + 6x^r(r^3 - 3r^2 + 2r) + 9x^r(r^2 - r) + 3x^{r} r + x^r = 0 \] \[ x^r \left((r^4 - 6r^3 + 11r^2 - 6r) + 6(r^3 - 3r^2 + 2r) + 9(r^2 - r) + 3r + 1\right) = 0 \] \[ x^r \left(r^4 - 6r^3 + 11r^2 - 6r + 6r^3 - 18r^2 + 12r + 9r^2 - 9r + 3r + 1\right) = 0 \] \[ x^r \left(r^4 - 6r^3 + 6r^3 + 11r^2 - 18r^2 + 9r^2 - 6r + 12r - 9r + 3r + 1\right) = 0 \] \[ x^r (r^4 + 2r^2 + 1) = 0 \] \[ x^r (r^2 + 1)^2 = 0 \]

Therefore the roots to the equation are: \( r = \pm i, \pm i. \) Hence the solution is:

\[ y(x) = C_1 x^0 \cos(\ln(x)) + C_2 x^0 \sin(\ln(x)) + C_3 x^0 \ln(x) \cos(\ln(x)) + C_4 x^0 \ln(x) \sin(\ln(x)) \] \[ y(x) = C_1 \cos(\ln(x)) + C_2 \sin(\ln(x)) + C_3 \ln(x) \cos(\ln(x)) + C_4 \ln(x) \sin(\ln(x)) \]

Practice

Evaluate these expressions:

  1. \( x^4 y^{(4)} + 8x^3 y^{(3)} + 15x^2 y'' + 8xy' + y = 0 \)
  2. \( 2x^2 y'' + 4xy' + y = 0 \)
  3. \( x^3 y''' + xy' - y = 0 \)

Solution:

  1. \( y(x) = C_1 x^{-1} + C_2 x^{-1} \ln(x) + C_3 \cos(\ln(x)) + C_4 \sin(\ln(x)) \)
  2. \( y(x) = C_1 x^{-\frac{1}{2}} \cos\left(\frac{1}{2} \ln(x)\right) + C_2 x^{-\frac{1}{2}} \sin\left(\frac{1}{2} \ln(x)\right) \)
  3. \( y(x) = C_1 x + C_2 x \ln(x) + C_3 x \ln^2(x) \)