Lesson 15 (09/28/26): Homogeneous equations with constant coefficients (3.3) - I

Lesson 15 (09/28/26): Homogeneous equations with constant coefficients (3.3) - I

Review: Solving \( ay'' + by' + cy = 0 \)

• Want two L.I. solutions \( y_1, y_2 \rightsquigarrow y = c_1 y_1 + c_2 y_2 \) is general sol.
• Look for \( y = e^{rx} \) as a solution: \[ y' = r e^{rx}, \quad y'' = r^2 e^{rx} \] \[ ay'' + by' + cy = 0 \rightsquigarrow \underbrace{e^{rx}}_{\neq 0} [\underbrace{ar^2 + br + c}_{= 0 \text{ is ch. equations}}] = 0 \]

Cases for Roots:

• Two distinct real roots \( r_1, r_2 \):

  \( y_1 = e^{r_1 x}, \quad y_2 = e^{r_2 x} \) are two L.I. solutions

  \( y = c_1 e^{r_1 x} + c_2 e^{r_2 x} \) is gen. sol.

• Only one real root \( r \):

  \( y_1 = e^{rx}, \quad y_2 = x e^{rx} \) are two L.I. solutions

• C'x Roots:

  Next Lecture

Example: \( y'' + 5y' - 24y = 0 \)

Ch. equation: \( r^2 + 5r - 24 = 0 \)

Quadratic formula

\[ r = \frac{-5 \pm \sqrt{25 + 96}}{2} \] \[ = \frac{-5 \pm 11}{2} \]

\( r_1 = 3 \), \( r_2 = -8 \)

Two distinct Real Roots

\( y_1 = e^{3x} \), \( y_2 = e^{-8x} \) are two L.I Solutions

\( y = c_1 y_1 + c_2 y_2 = c_1 e^{3x} + c_2 e^{-8x} \) is the general solution

eg: \( y'' - 10y' + 25y = 0 \)

ch. equation:

\[ r^2 - 10r + 25 = 0 \] \[ (r - 5)^2 = 0 \]

just one real root \( r = 5 \) is Repeating twice give us two L.I solutions

\[ y_1 = e^{5x}, \quad y_2 = x e^{5x} \]

General solution:

\[ y = c_1 e^{5x} + c_2 x e^{5x} \]

\( n^{\text{th}} \) Order Linear Homogeneous ODE with Constant Coefficients

\[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_2 y'' + a_1 y' + a_0 y = 0 \]

Need \( n \) L.I. solutions \( y_1, \dots, y_n \) to write general solution \( y = c_1 y_1 + c_2 y_2 + \dots + c_n y_n \)

Idea: Look for \( y = e^{rx} \) as a solution

\[ y' = r e^{rx}, \quad y'' = r^2 e^{rx}, \quad y''' = r^3 e^{rx}, \dots, \quad y^{(n)} = r^n e^{rx} \]

Plug in to ODE:

\[ \underbrace{e^{rx}}_{\neq 0} \left[ a_n r^n + a_{n-1} r^{n-1} + \dots + a_2 r^2 + a_1 r + a_0 \right] = 0 \]

Find \( r \) such that

\[ a_n r^n + a_{n-1} r^{n-1} + \dots + a_1 r + a_0 = 0 \quad \Big\} \text{ Characteristic equation.} \]

Case (i) Distinct Real Roots

Ch. equation:

\[ a_n r^n + a_{n-1} r^{n-1} + \dots + a_2 r^2 + a_1 r + a_0 = 0 \]

Suppose \( r_1, r_2, r_3, \dots, r_n \) are distinct roots

then \( y_1 = e^{r_1 x} \), \( y_2 = e^{r_2 x} \), \( y_3 = e^{r_3 x} \), \dots, \( y_n = e^{r_n x} \)

are \( n \)- L.I solutions.

\[ W(y_1, \dots, y_n) = \begin{vmatrix} e^{r_1 x} & \dots & e^{r_n x} \\ r_1 e^{r_1 x} & & \vdots \\ \vdots & & \vdots \\ r_1^{n-1} e^{r_1 x} & \dots & r_n^{n-1} e^{r_n x} \end{vmatrix} \neq 0 \]

Gen sol: \( y = c_1 e^{r_1 x} + c_2 e^{r_2 x} + \dots + c_n e^{r_n x} \)

Example:

\[ y''' - 3y'' - 10y' = 0 \]

Ch. equation:

\[ r^3 - 3r^2 - 10r = 0 \] \[ r(r^2 - 3r - 10) = 0 \] \[ r(r - 5)(r + 2) = 0 \]

\( r = 0, \, r = -2, \, r = 5 \) are \(3\) distinct roots

\( y_1 = e^{0x} = 1, \, y_2 = e^{-2x}, \, y_3 = e^{5x} \) are L.I solutions.

\[ W(y_1, y_2, y_3) = \begin{vmatrix} 1 & e^{-2x} & e^{5x} \\ 0 & -2e^{-2x} & 5e^{5x} \\ 0 & 4e^{-2x} & 25e^{5x} \end{vmatrix} = 1 \begin{vmatrix} -2e^{-2x} & 5e^{5x} \\ 4e^{-2x} & 25e^{5x} \end{vmatrix} = -70 e^{3x} \neq 0 \]

General solution: \( y = c_1 + c_2 e^{-2x} + c_3 e^{5x} \)

Case (ii) Repeated Real Roots

eg: \( y'' - 10y' + 25 = 0 \rightsquigarrow \) ch. equation \( r^2 - 10r + 25 = (r - 5)^2 = 0 \)

\( r = 5 \) repeats twice
\( \Rightarrow \) corresponds to two L.I. solutions
\[ e^{5x}, \quad x e^{5x} \]

\( a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_1 y' + a_0 = 0 \rightsquigarrow \) ch. equation

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

Suppose \( (r - r_0)^k \) is a factor.
\( r = r_0 \) corresponds to \( k \) L.I. solutions of ODE

\[ y_1 = e^{r_0 x}, \quad y_2 = x e^{r_0 x}, \quad \dots, \quad y_k = x^{k-1} e^{r_0 x} \]

Eg: \( y''' - 3y'' + 3y' - y = 0 \)

Ch. equation: \( r^3 - 3r^2 + 3r - 1 = 0 \)

\[ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \] \[ \Rightarrow (r-1)^3 = 0 \]

\( r=1 \) is root repeating 3 times

\( y_1 = e^x \), \( y_2 = xe^x \), \( y_3 = x^2e^x \) are 3 L.I solutions.

\[ W(y_1, y_2, y_3) = \begin{vmatrix} e^x & xe^x & x^2 e^x \\ e^x & xe^x + e^x & x^2 e^x + 2xe^x \\ e^x & xe^x + 2e^x & x^2 e^x + 4xe^x + 2e^x \end{vmatrix} \] \[ \neq 0 \]

\( y_2 = x e^x \) is a solution

\[ y_2''' - 3y_2'' + 3y_2' - 1 = \cancel{x e^x} + \cancel{3e^x} - \cancel{3x e^x} - \cancel{6e^x} + \cancel{3x e^x} + \cancel{3e^x} - \cancel{x e^x} \] \[ \begin{aligned} y_2' &= x e^x + e^x \\ y_2'' &= x e^x + 2e^x \\ y_2''' &= x e^x + 3e^x \end{aligned} \] \[ = 0 \]

\( y_3 = x^2 e^x \) is a solution.

\[ y_3' = x^2 e^x + 2x e^x, \quad y_3'' = x^2 e^x + 4x e^x + 2e^x \] \[ \begin{aligned} y_3''' &= x^2 e^x + 2x e^x + 4x e^x + 4e^x + 2e^x \\ &= x^2 e^x + 6x e^x + 6e^x \end{aligned} \] \[ y_3''' - 3y_3'' + 3y_3' - 1 = \cancel{x^2 e^x} + \cancel{6x e^x} + \cancel{6e^x} - \cancel{3x^2 e^x} - \cancel{12x e^x} - \cancel{6e^x} + \cancel{3x^2 e^x} + \cancel{6x e^x} - \cancel{x^2 e^x} = 0. \]

eg: \( y^{(5)} - 6y^{(4)} + 9y^{(3)} = 0 \)

ch. equation:

\[ r^5 - 6r^4 + 9r^3 = 0 \] \[ r^3 [r^2 - 6r + 9] = 0 \] \[ r^3 (r - 3)^2 = 0 \]

\( r = 0 \) is repeats 3 times \( \Rightarrow \) 3 L.I solutions

\[ e^{0x}, \; x e^{0x}, \; x^2 e^{0x} \]

\( r = 3 \) is repeats 2 times \( \Rightarrow \) 2 L.I solutions

\[ e^{3x}, \; x e^{3x} \]

General solution: \( y = c_1 + c_2 x + c_3 x^2 + c_4 e^{3x} + c_5 x e^{3x} \)

eg: Suppose ch. equation of a 8th order ODE is

\[ (r-1)^4 (r+1)^2 (r-3)(r+5) = 0 \]

find the gen. solution

\( r=1 \leadsto \) repeats 4 times \( \leadsto \) 4 L.I solutions

\[ e^x, \; x e^x, \; x^2 e^x, \; x^3 e^x \]

\( r=-1 \leadsto \) repeats 2 times \( \leadsto \) 2 L.I solutions

\[ e^{-x}, \; x e^{-x} \]

\( r=3, \; r=-5 \leadsto \) distinct roots \( \leadsto \) 2 L.I solutions

\[ e^{3x}, \; e^{-5x} \]

Gen. sol:

\[ y = c_1 e^x + c_2 x e^x + c_3 x^2 e^x + c_4 x^3 e^x + c_5 e^{-x} + c_6 x e^{-x} + c_7 e^{3x} + c_8 e^{-5x}. \]