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5.3 Homogeneous Eqs. w/ Constant Coefficients (part 2)

\[ y'' + ay' + by = 0 \quad a, b \text{ constants} \]

solutions: \( y = e^{rx} \)

\( r \): solutions of characteristic eq.

\[ r^2 + ar + b = 0 \]

two cases:

  • \( r \)'s are real and distinct \( \rightarrow r_1, r_2 \quad r_1 \neq r_2 \)

    \[ y_1 = e^{r_1 x} \quad y_2 = e^{r_2 x} \]

  • \( r \)'s are repeated \( \rightarrow r_1 = r_2 = r \)

    \[ y_1 = e^{rx} \quad y_2 = xe^{rx} \]

today: \( r \)'s are complex

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start w/: \( y'' + y = 0 \)

characteristic eq \( r^2 + 1 = 0 \)

\[ r^2 = -1 \]

\[ r = \pm \sqrt{-1} = \pm i \]

\[ i^2 = -1 \]

solutions: \( e^{rx} \)

\( \rightarrow e^{ix}, e^{-ix} \quad e^{ix} = ? \)

\[ y'' + y = 0 \rightarrow y'' = -y \]

solution is a function that is the negative of its 2nd derivative

exponential? \( y = e^{ax} \quad y' = ae^{ax} \quad y'' = a^2 e^{ax} \)

same sign!

so, NOT exponential

but \( y = \cos x, y' = -\sin x, y'' = -\cos x \)

\( y = \sin x, y' = \cos x, y'' = -\sin x \)

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Euler's Formula and Taylor Series

So, \( e^{ix} \) and \( e^{-ix} \) must be related to \( \cos x \) and \( \sin x \).

How?

\( e^{ix} = ? \)

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \quad \text{(Taylor series of } e^x \text{)} \]

Now replace \( x \) with \( ix \):

\[ e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots \]

Powers of \( i \):

  • \( i = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)
  • \( i^4 = 1 \)
  • \( i^5 = i \)
  • \( i^6 = -1 \)
  • \( i^7 = -i \)
  • \( i^8 = 1 \)

cycle repeats

\[ = 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} + i \frac{x^5}{5!} + \dots \]
\[ = \underbrace{\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} + \dots \right)}_{\cos x} + i \underbrace{\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} + \dots \right)}_{\sin x} \]

\( e^{ix} = \cos x + i \sin x \)

Euler's Formula

\( e^{-ix} = \cos x - i \sin x \)

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Example: Solving Differential Equations

Example: \( y'' + 100y = 0 \)

\[ r^2 + 100 = 0 \implies r = \pm \sqrt{-100} = \pm i \sqrt{100} = \pm 10i \]

Solutions:

\( e^{10ix} = e^{i(10x)} = \cos(10x) + i \sin(10x) \) — ①

\( e^{-10ix} = e^{i(-10x)} = \cos(-10x) + i \sin(-10x) \)

\( = \cos(10x) - i \sin(10x) \) — ②

These solutions are complex but we know solutions should be real-valued.

① + ② divided by 2 \( \to \cos(10x) \)

① - ② divided by \( 2i \to \sin(10x) \)

Simply the real and imaginary parts of \( e^{rx} \)

These are taken to be the two solutions (real-valued) that span the solution space.

So, \( y_1 = \cos(10x) \quad y_2 = \sin(10x) \)

General solution: \( y = C_1 \cos(10x) + C_2 \sin(10x) \)

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So, the solutions are just the real and imaginary parts of \( e^{rx} \) w/ \( r \) being complex.

Example

\[ y'' - 6y' + 25y = 0 \]
\[ r^2 - 6r + 25 = 0 \quad \text{cannot factor this} \]
\[ r = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(25)}}{2(1)} \]
\[ = \frac{6 \pm \sqrt{-64}}{2} = \frac{6 \pm 8i}{2} = 3 \pm 4i \]

form one solution, grab the real and imaginary parts

use \( r = 3 + i4 \)

\[ e^{rx} = e^{(3+4i)x} = e^{3x} e^{4ix} = e^{3x} e^{i(4x)} \]
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\[ e^{ix} = \cos x + i \sin x \]
\[ e^{3x} e^{i(4x)} = e^{3x} (\cos 4x + i \sin 4x) \]
\[ = (e^{3x} \cos 4x) + i(e^{3x} \sin 4x) \]
\( e^{3x} \cos 4x \)
real part
\( e^{3x} \sin 4x \)
imaginary part

use the real and imag parts to span solution space

\( y_1 = e^{3x} \cos 4x \)

\( y_2 = e^{3x} \sin 4x \)

General solution:

\[ y = c_1 e^{3x} \cos 4x + c_2 e^{3x} \sin 4x \]
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Example: Solving a Fourth-Order Differential Equation

Given the differential equation:

\[6y^{(4)} + 11y'' + 4y = 0\]

We find the characteristic equation:

\[6r^4 + 11r^2 + 4 = 0\]

This can be rewritten as a quadratic in terms of \(r^2\):

\[6(r^2)^2 + 11(r^2) + 4 = 0\]

Let \(u = r^2\). Then the equation becomes:

\[6u^2 + 11u + 4 = 0\]

Using the quadratic formula to solve for \(u\):

\[u = \frac{-11 \pm \sqrt{11^2 - 4(6)(4)}}{2(6)} = \frac{-11 \pm \sqrt{25}}{12}\]
\[u = \frac{-11 \pm 5}{12} = -\frac{4}{3}, -\frac{1}{2}\]

Since \(u = r^2\), we have:

\[r^2 = -\frac{4}{3}, \quad r^2 = -\frac{1}{2}\]

Solving for \(r\) gives the four roots:

\[r = \frac{2}{\sqrt{3}}i, -\frac{2}{\sqrt{3}}i, \frac{1}{\sqrt{2}}i, -\frac{1}{\sqrt{2}}i\]
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Constructing the Solution

Form one solution using \(r = \pm \frac{2}{\sqrt{3}}i\) and grab the real and imaginary parts:

\[e^{\frac{2}{\sqrt{3}}ix} = \left[ \cos\left(\frac{2}{\sqrt{3}}x\right) \right] + i \left[ \sin\left(\frac{2}{\sqrt{3}}x\right) \right]\]

\(y_1 = \cos\left(\frac{2}{\sqrt{3}}x\right)\)

\(y_2 = \sin\left(\frac{2}{\sqrt{3}}x\right)\)

Repeat with \(r = \pm \frac{1}{\sqrt{2}}i\):

\[e^{\frac{1}{\sqrt{2}}ix} = \left[ \cos\left(\frac{1}{\sqrt{2}}x\right) \right] + i \left[ \sin\left(\frac{1}{\sqrt{2}}x\right) \right]\]

\(y_3 = \cos\left(\frac{1}{\sqrt{2}}x\right)\)

\(y_4 = \sin\left(\frac{1}{\sqrt{2}}x\right)\)

General Solution

The general solution is given by \(y = c_1 y_1 + c_2 y_2 + c_3 y_3 + c_4 y_4\):

\[y = c_1 \cos\left(\frac{2}{\sqrt{3}}x\right) + c_2 \sin\left(\frac{2}{\sqrt{3}}x\right) + c_3 \cos\left(\frac{1}{\sqrt{2}}x\right) + c_4 \sin\left(\frac{1}{\sqrt{2}}x\right)\]
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Example: Solving a Third-Order Linear Homogeneous Differential Equation

Consider the following third-order linear homogeneous differential equation with constant coefficients:

\[ y''' - 5y'' + 100y' - 500y = 0 \]

Step 1: Characteristic Equation

We first write the characteristic equation by substituting \( y = e^{rx} \):

\[ r^3 - 5r^2 + 100r - 500 = 0 \]

Step 2: Factoring by Grouping

Group the terms to factor the cubic polynomial:

\[ r^2(r - 5) + 100(r - 5) = 0 \] \[ (r - 5)(r^2 + 100) = 0 \]

Step 3: Finding the Roots

Setting each factor to zero gives the roots of the characteristic equation:

\[ r_1 = 5, \quad r_2 = 10i, \quad r_3 = -10i \]

Note: Complex roots always come in conjugate pairs.

Step 4: General Solution

Using the roots found above, the general solution to the differential equation is:

\[ y = c_1 e^{5x} + c_2 \cos(10x) + c_3 \sin(10x) \]