Lesson 16 (09/30/26): Homogeneous Equations with Constant Coefficients (3.3) - II

Lesson 16 (09/30/26): Homogeneous Equations with Constant Coefficients (3.3) - II

Example: Find general solution to \( y''' + 3y'' - 4y = 0 \)

Ch. equation: \[ r^3 + 3r^2 - 4 = 0 \]

Guess: \( r = 1 \) is a root, \( r - 1 \) is a factor

Find other factors using long division:

\[ \begin{array}{r} r^2 + 4r + 4\phantom{)} \\ r - 1{\overline{\smash{\big)}\,r^3 + 3r^2\phantom{+ 0r} - 4\phantom{)}}} \\ \underline{-(r^3 - \phantom{3}r^2)\phantom{-0r-4)}} \\ 4r^2\phantom{+ 0r} - 4\phantom{)} \\ \underline{-(4r^2 - 4r)\phantom{-4)}} \\ 4r - 4\phantom{)} \\ \underline{-(4r - 4)} \\ 0\phantom{)} \end{array} \]

\[ \begin{aligned} r^3 + 3r^2 - 4 &= (r - 1)(r^2 + 4r + 4) \\ &= (r - 1)(r + 2)^2 \end{aligned} \]

\( r = 1 \rightsquigarrow \text{one solution: } e^x \)

\( r = -2 \rightsquigarrow \text{two solutions: } e^{-2x},\; x e^{-2x} \)

General Solution: \[ y = c_1 e^x + c_2 e^{-2x} + c_3 x e^{-2x} \]

Solve: \( y'' + 4y = 0 \)

Ch. equation is

\[ r^2 + 4 = 0 \]

Define: \( \sqrt{-1} = i \)

\( r = \pm 2i \) are two roots

Consider: \( y = e^{2ix} \leadsto y' = (2i)e^{2ix} \leadsto y'' = (2i)^2 e^{2ix} = -4e^{2ix} = -4y \)

Similarly \( y = e^{-2ix} \leadsto y'' = -4y \)

\( e^{2ix} \) & \( e^{-2ix} \) are two complex solutions.

Euler formula:

\[ e^{i\theta} = \cos\theta + i\sin\theta \] \[ e^{2ix} = \cos 2x + i\sin 2x \] \[ e^{-2ix} = \cos 2x - i\sin 2x \]

\( y_1 = \cos 2x \leadsto y_1' = -2\sin 2x \), \( y_1'' = -4\cos 2x = -4y_1 \)

\( y_2 = \sin 2x \leadsto y_2' = 2\cos 2x \), \( y_2'' = -4\sin 2x = -4y_2 \)

\( y_1 = \cos 2x \) & \( y_2 = \sin 2x \) are two Real Solutions.

Gen. solution: \( y = C_1 \cos 2x + C_2 \sin 2x \)

Suppose \( Y(x) = Y_R(x) + i\, Y_I(x) \) is a complex solution to Homogeneous, linear ODE then \( Y_R(x) \) & \( Y_I(x) \) are two Real solutions.

\( n^{\text{th}} \) derivative = \( Y^{(n)}(x) = Y_R^{(n)}(x) + i\, Y_I^{(n)}(x) \) for any \( n \).

\[ a_n Y^{(n)}(x) + a_{n-1} Y^{(n-1)}(x) + \dots + a_2 Y''(x) + a_1 Y'(x) + a_0 Y(x) = 0 \] \[ a_n \left[ Y_R^{(n)} + i\, Y_I^{(n)} \right] + a_{n-1} \left[ Y_R^{(n-1)}(x) + i\, Y_I^{(n-1)}(x) \right] + \dots + a_0 \left[ Y_R(x) + i\, Y_I(x) \right] = 0 \] \[ \left( a_n Y_R^{(n)} + \dots + a_1 Y_R' + a_0 Y_R \right) + i \left( a_n Y_I^{(n)} + \dots + a_1 Y_I' + a_0 Y_I \right) = 0 \] \[ \left.\begin{aligned} \left( a_n Y_R^{(n)} + \dots + a_1 Y_R' + a_0 Y_R \right) &= 0 \\[6pt] \left( a_n Y_I^{(n)} + \dots + a_1 Y_I' + a_0 Y_I \right) &= 0 \end{aligned}\right\} \quad Y_R \text{ \& } Y_I \text{ are two Real solutions.} \]

\[ ay'' + by' + cy = 0 \]

Ch. equation: \( ar^2 + br + c = 0 \)

Suppose \( r = \alpha \pm i\beta \) are two c'x roots.

\[ e^{rx} = e^{\alpha x \pm i\beta x} = e^{\alpha x} [e^{\pm i\beta x}] \]

use Euler

\[ = e^{\alpha x} [\cos\beta x \pm i \sin\beta x] \] \[ = e^{\alpha x} \cos\beta x \pm i e^{\alpha x} \sin\beta x \] \[ y_1 = e^{\alpha x} \cos\beta x \] \[ y_2 = e^{\alpha x} \sin\beta x \]

are two Real solutions

Gen. solution: \( y = c_1 e^{\alpha x} \cos\beta x + c_2 e^{\alpha x} \sin\beta x \)

Example

\[ y'' + 6y' + 13y = 0 \]

Ch. equation: \[ r^2 + 6r + 13 = 0 \]

\[ \text{roots} = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm 4i}{2} = -3 \pm 2i \]

\[ e^{(-3+2i)x} = e^{-3x} \cdot \underbrace{e^{2ix}}_{\text{Euler}} \]

\[ = e^{-3x} [\cos 2x + i \sin 2x] = (e^{-3x} \cos 2x) + i (e^{-3x} \sin 2x) \]

\[ y_1 = e^{-3x} \cos 2x, \quad y_2 = e^{-3x} \sin 2x \]

are two Real solutions

General solution: \[ y = c_1 e^{-3x} \cos 2x + c_2 e^{-3x} \sin 2x \]

eg: \[ y^{(4)} + 8y'' + 16y = 0 \]

ch:

\[ r^4 + 8r^2 + 16 = 0 \] \[ (r^2 + 4)^2 = 0 \]

\( r = 2i, -2i \) are two complex roots repeating twice

Already know \( \rightsquigarrow e^{2ix} = \cos 2x + i \sin 2x \)

gives 2 Real Roots \( \cos 2x, \sin 2x \).

Just as we did in the Repeated Real root case multiply with \(x\) to get the remaining solutions

4 solutions: \( \cos 2x, \sin 2x, x \cos 2x, x \sin 2x \)

Gen. solution: \( y = c_1 \cos 2x + c_2 \sin 2x + c_3 x \cos 2x + c_4 x \sin 2x \)

Let's verify \( x \cos 2x \) is a solution

\[ y = x \cos 2x \] \[ y' = \cos 2x - 2x \sin 2x \] \[ y'' = -2 \sin 2x - 2 \sin 2x - 4x \cos 2x \] \[ y''' = -8 \cos 2x - 4 \cos 2x + 8x \sin 2x \] \[ y^{(4)} = 24 \sin(2x) + 8 \sin 2x + 16x \cos 2x \] \[ \begin{aligned} y^{(4)} + 8y'' + 16y &= 32 \sin(2x) + 16x \cos 2x \\ &\quad - 32 \sin(2x) - 32x \cos 2x \\ &\quad + 16x \cos 2x = 0 \end{aligned} \]

Eg: Ch. equation of ODE is \((r^2+4)^3 = 0\)

\(\pm 2i\) are roots repeated \(3\) times

\(\Rightarrow \cos 2x, \sin 2x\) are solutions

Keep multiplying each with \(x\) untill you get total \(6\) solutions

\(\Rightarrow 6\) solutions:

\[ \begin{aligned} &\cos 2x, \sin 2x \\ &x \cos 2x, x \sin 2x \\ &x^2 \cos 2x, x^2 \sin 2x \end{aligned} \]

Suppose \(9^{\text{th}}\) order ODE has ch. equation

\[ (r+3)(r-2)^2(r^2+16)(r^2-10r+26)^2 = 0 \]

find general solutions

\(r+3=0 \rightsquigarrow r=-3\) No repetition \(\rightsquigarrow\) gives 1 solution \(e^{-3x}\)

\((r-2)^2=0 \rightsquigarrow r=2\) Repeating twice \(\rightsquigarrow\) gives 2 solutions \(e^{2x}, xe^{2x}\)

\(r^2+16=0 \rightsquigarrow r=\pm 4i\) Non Repeating cx Roots \(\rightsquigarrow\) two solutions \(\cos(4x), \sin(4x)\)

\((r^2-10r+26)^2=0 \rightsquigarrow r=5\pm i\) Repeating twice cx Roots \(\Rightarrow 4\) solutions

\[ e^{5x}\cos x, \quad e^{5x}\sin x \] \[ xe^{5x}\cos x, \quad xe^{5x}\sin x \]

Gen. sol:

\[ y = c_1 e^{-3x} + c_2 e^{2x} + c_3 x e^{2x} + c_4 \cos(4x) + c_5 \sin(4x) + c_6 e^{5x}\cos x + c_7 e^{5x}\sin x + c_8 x e^{5x}\cos x + c_9 x e^{5x}\sin x \]

eg: \( y^{(6)} + 4y^{(4)} + 4y'' = 0 \)

ch. equation:

\[ r^6 + 4r^4 + 4r^2 = 0 \] \[ r^2 [r^4 + 4r^2 + 4] = 0 \] \[ r^2 [r^2 + 2]^2 = 0 \]

\( r^2 = 0 \rightsquigarrow \) two solutions \( \left. e^{0x}, x e^{0x} \right\} 1, x \).

\( [r^2 + 2]^2 = 0 \rightsquigarrow 4 \) solutions

roots \( r = \pm \sqrt{2}i \)

\[ \cos\sqrt{2}x, \quad \sin\sqrt{2}x \]

\[ x\cos\sqrt{2}x, \quad x\sin\sqrt{2}x. \]