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5.3 Homogeneous Eqs. with Constant Coefficients

homogeneous: right side zero

\[ y'' + 5y' - 2y = 0 \]\[ y''' + 10y'' - 5y' + 17 = 0 \]\[ y^{(10)} - 100y = 0 \]

find a \( y \) that is a constant multiple of its derivatives

\[ y = e^{rx} \]

\( y' = re^{rx} \)

\( r \) times itself

if we substitute \( y = e^{rx} \) (because it solves those eqs.) into them, we get the characteristic eq. that we solve to find \( r \).

for example,

\[ 2y'' - 3y' = 0 \]

\[ y = e^{rx} \]\[ y' = re^{rx} \]\[ y'' = r^2 e^{rx} \]

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\[ 2r^2 e^{rx} - 3re^{rx} = 0 \]\[ e^{rx}(2r^2 - 3r) = 0 \quad e^{rx} \neq 0 \]

\[ 2r^2 - 3r = 0 \]

Characteristic eq.

we can get this directly by using coefficients of the differential eq. \( \rightarrow \)

coeff of \( y'' \) becomes coeff of \( r^2 \)
coeff of \( y' \) becomes coeff of \( r \)
coeff of \( y \) becomes coeff of \( 1 \)

\[ 2r^2 - 3r = 0 \]\[ r(2r - 3) = 0 \rightarrow r = 0, \quad r = 3/2 \quad \text{distinct roots} \]

Solutions:

\[ y_1 = e^{0x} = 1 \]\[ y_2 = e^{3/2 x} \]

General solution: linear combo of them

\[ y = c_1 \cdot 1 + c_2 e^{3/2 x} \]

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Example: Second-Order Linear Homogeneous ODE

\[ 4y'' - 12y' + 9y = 0 \]

Mapping coefficients to the characteristic equation:

The coefficient of \( y'' \) (4) becomes the coefficient of \( r^2 \).
The coefficient of \( y' \) (-12) becomes the coefficient of \( r \).
The coefficient of \( y \) (9) becomes the constant term (coefficient of 1).

Characteristic Equation

\[ 4r^2 - 12r + 9 = 0 \]

Factoring the quadratic:

\[ (2r - 3)(2r - 3) = 0 \]

Solving for \( r \):

\[ r = 3/2, \, 3/2 \quad \text{(repeated)} \]

Fundamental Solutions

\[ y_1 = e^{3/2 x} \]\[ y_2 = x e^{3/2 x} \]

Note: Multiply by \( x \) each time a root is repeated.

General Solution

\[ y = c_1 e^{3/2 x} + c_2 x e^{3/2 x} \]

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Higher-Order Repeated Roots

\[ y''' = 0 \]

Characteristic equation:

\[ r^3 = 0 \implies r = 0, 0, 0 \]

Individual solutions:

\[ y_1 = e^{0x} = 1 \]\[ y_2 = x e^{0x} = x \]\[ y_3 = x^2 e^{0x} = x^2 \]

Rule: For each repeat, one factor of \( x \) is multiplied.

General solution:

\[ y = c_1 + c_2 x + c_3 x^2 \]

Example: Fourth-Order ODE

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

Characteristic equation:

\[ r^4 - 8r^3 + 16r^2 = 0 \]

Factoring:

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

Roots:

\[ r = 0, 0, 4, 4 \]

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\[ \begin{aligned} y_1 &= e^{0x} = 1 \\ y_2 &= x e^{0x} = x \end{aligned} \]

\}

from \( r = 0, 0 \)

\[ \begin{aligned} y_3 &= e^{4x} \\ y_4 &= x e^{4x} \end{aligned} \]

\}

from \( r = 4, 4 \)

Gen. solution: \( y = c_1 + c_2 x + c_3 e^{4x} + c_4 x e^{4x} \)

Working Backwards

We can work backwards:

Solution is \( y = c_1 e^{-2x} + c_2 x e^{-2x} + c_3 x^2 e^{-2x} \)

What is the differential eq?

\( y_1 = e^{-2x} \) \(\rightarrow\) \( r = -2 \)
\( y_2 = x e^{-2x} \) \(\rightarrow\) repeat of \( r = -2 \) (\( x \) in front \(\rightarrow\) repeat)
\( y_3 = x^2 e^{-2x} \) \(\rightarrow\) \( r = -2 \) (2nd repeat)

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Characteristic Equation

\( (r+2)(r+2)(r+2) = 0 \)

\( (r+2)(r^2 + 4r + 4) = 0 \)

\( r^3 + 6r^2 + 12r + 8 = 0 \)

\(\downarrow\)

coeff of \( y''' \)

\(\downarrow\)

coeff of \( y'' \)

Differential Equation

Differential eq: \( y''' + 6y'' + 12y' + 8y = 0 \)

Cubic characteristic eqs can be hard to solve (no nice formula like quadratic formula).

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Example: Third-Order Linear Homogeneous ODE

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

Characteristic equation:

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

Here, there is a pattern in coefficients: 1, 5, -100, -500.

The first pair (1, 5) shows a ratio where 5 is a multiple of 1.
The second pair (-100, -500) shows a ratio where -500 is a multiple of -100 (specifically, multiplied by 5).

Solving by Grouping

\[ r^3 + 5r^2 - 100r - 500 = 0 \]\[ r^2(r + 5) - 100(r + 5) = 0 \]\[ (r + 5)(r^2 - 100) = 0 \]\[ (r + 5)(r + 10)(r - 10) = 0 \]

The roots are:

\[ r = -5, -10, 10 \]

The general solution is:

\[ y = c_1 e^{-5x} + c_2 e^{-10x} + c_3 e^{10x} \]

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Example: Factoring Higher-Order Equations

\[ 3y''' + 4y'' - 5y' - 2y = 0 \]

Characteristic equation:

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

Coefficients: 3, 4, -5, -2. No nice pattern like last time.

Strategy: Finding a Root

Turn the cubic equation into the form:

\[ (r - a)(\text{quadratic}) = 0 \]

If we can find one root (\(a\)), then we just need to find that quadratic to proceed further.

We find the first root by inspection or trial and error.

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

One root comes from the constant term (-2):

\[ -2 = -2 \times 1 = -1 \times 2 \]

Possible roots to test: -2, 1, -1, or 2.

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Finding Roots of a Cubic Equation

One of these is a root: \( \pm 2, \pm 1 \)

Pick one and try: \( r = 1 \)

\( 3r^3 + 4r^2 - 5r - 2 = 0 \) is \( r = 1 \) a solution?

\( 3(1)^3 + 4(1)^2 - 5(1) - 2 = 0 \) so \( r = 1 \) is a root

\[ (r - 1)(ar^2 + br + c) = 0 \]

\( r = 1 \) from trial and error

?

Equating the expressions:

\[ (r - 1)(ar^2 + br + c) = 3r^3 + 4r^2 - 5r - 2 \]

Solving for the quadratic factor:

\[ ar^2 + br + c = \frac{3r^3 + 4r^2 - 5r - 2}{r - 1} \]

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Polynomial Long Division

\( 3r^2 + 7r + 2 \)

that quadratic
\( r - 1 \) \( 3r^3 + 4r^2 - 5r - 2 \)

\( - (3r^3 - 3r^2) \)

\( 7r^2 - 5r - 2 \)

\( - (7r^2 - 7r) \)

\( 2r - 2 \)

\( - (2r - 2) \)

0

Factoring and General Solution

\( (r - 1)(3r^2 + 7r + 2) = 0 \)

\( (r - 1)(3r + 1)(r + 2) = 0 \) \( r = 1, -1/3, -2 \)

\[ y = c_1 e^x + c_2 e^{-1/3 x} + c_3 e^{-2x} \]

Next time: complex roots \( \rightarrow e^{ix} = ? \)