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3.5 Undetermined Coeff. (continued)

\[ ay'' + by' + cy = f(t) \]
\[ y(t) = c_1 y_1 + c_2 y_2 + Y(t) \]

Complementary solution

from \( ay'' + by' + cy = 0 \)

particular solution

due to \( f(t) \)

Undetermined coeff: \( Y(t) \) resembles \( f(t) \)

  • polynomial \( f(t) \) → \( Y(t) \) is polynomial
  • exponential \( f(t) \) → \( Y(t) \) is exponential
  • cosine or sine \( f(t) \) → \( Y(t) \) has BOTH cosine and sine

What happens if \( f(t) \) copies one of the fundamental solutions?

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Example

For example, \[ y'' + y' - 2y = e^t \]

\[ y'' + y' - 2y = 0 \quad \implies \quad r^2 + r - 2 = 0 \quad \implies \quad (r+2)(r-1) = 0 \] \[ r = -2, \quad r = 1 \]
\[ y = c_1 e^{-2t} + c_2 e^t + Y \]

\( f(t) \) is \( e^t \), so normally we say \( Y = Ae^t \)

\[ Y' = Ae^t \] \[ Y'' = Ae^t \]

Sub into \( y'' + y' - 2y = e^t \)

\[ Ae^t + Ae^t - 2Ae^t = e^t \] \[ 0 = e^t \quad \text{not true!} \]

This says \( Y = Ae^t \) is NOT right because \( f(t) \) copies \( y_1 \) or \( y_2 \).

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Reduction of Order for Non-Homogeneous ODEs

Let's use the reduction of order to see what \( Y \) should look like.

\[ y'' + y' - 2y = e^t \]

Homogeneous solutions: \( y_1 = e^{-2t}, \quad y_2 = e^t \)

Reduction of Order

If we have one solution, we can use it to find the rest.

\[ y = v(t)y_1 \quad \text{or} \quad y = v(t)y_2 \]

Here, let's use \( y_2 = e^t \).

\[ y = v e^t \quad \text{find } v \]
\[ y' = v e^t + v' e^t \]
\[ y'' = v e^t + v' e^t + v' e^t + v'' e^t = v e^t + 2v' e^t + v'' e^t \]

Sub into \( y'' + y' - 2y = e^t \):

\[ v e^t + 2v' e^t + v'' e^t + v e^t + v' e^t - 2v e^t = e^t \]
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Divide by \( e^t \):

\[ v'' + 3v' = 1 \]

Let \( w = v' \):

\[ w' + 3w = 1 \quad \text{(linear 1st-order)} \]
\[ \mu = e^{\int 3 dt} = e^{3t} \]
\[ \frac{d}{dt}(e^{3t} w) = e^{3t} \]
\[ e^{3t} w = \frac{1}{3} e^{3t} + a \]
\[ v' = w = \frac{1}{3} + a e^{-3t} \]
\[ v = \frac{1}{3} t - \frac{1}{3} a e^{-3t} + b \quad \text{where } y = v e^t \]
\[ y = -\frac{1}{3} a e^{-3t} \cdot e^t + b \cdot e^t + \frac{1}{3} t \cdot e^t \]
\[ y = c_1 e^{-2t} + c_2 e^t + \frac{1}{3} t e^t \]
Complementary solution: \( c_1 e^{-2t} + c_2 e^t \)
Particular form is \( t e^t \)

Attach an extra \( t \) if \( f(t) \) copies \( y_1 \) or \( y_2 \).

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Example: Solving a Non-Homogeneous Differential Equation

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

The general solution is of the form:

\[ y = c_1 y_1 + c_2 y_2 + Y \]

Step 1: Solve the Homogeneous Equation

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

Characteristic equation:

\[ r^2 - r = 0 \implies r = 0, r = 1 \]

Fundamental solutions:

\[ y_1 = e^{0t} = 1, \quad y_2 = e^t \]

So the complementary solution is:

\[ y = c_1 + c_2 e^t + Y \]

Step 2: Find the Particular Solution \( Y \)

Given \( f(t) = 3 \) (a constant).

So, normally we say \( Y = A \), but \( A = A \cdot 1 \) which copies \( c_1 = c_1 \cdot 1 \). We must attach \( t \) to the \( Y \) that copies \( y_1 \) or \( y_2 \).

\[ Y = At \]\[ Y' = A \]\[ Y'' = 0 \]

Substitute into the original equation:

\[ -A = 3 \implies A = -3 \]

Final Solution

\[ y = c_1 + c_2 e^t - 3t \]
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Example: Polynomial Non-Homogeneous Term

\[ y'' - y' = t \]

General form: \( y = c_1 + c_2 e^t + Y \)

\( f(t) = t \) is a 1st-order polynomial.

Guess the form: \( Y = At + B \)

Note: The constant term \( B \) copies \( c_1 \). Give it a \( t \), but it will then copy \( At \), so we give BOTH a \( t \).

Fix:

\[ Y = At^2 + Bt \]\[ \text{or } Y = t(At + B) \]

Example: Combined Terms

\[ y'' - y' = t + e^{2t} \]

\( y = c_1 + c_2 e^t + Y \)

Initial guess for \( Y \):

\[ Y = At + B + Ce^{2t} \]

Analysis:

  • For \( At + B \): Needs an extra \( t \) for copying \( y_1 \) or \( y_2 \).
  • For \( Ce^{2t} \): No \( t \) because \( e^{2t} \) is not copying anything.

Fixed \( Y \) is:

\[ Y = t(At + B) + Ce^{2t} \]
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Example: Method of Undetermined Coefficients

\[ y'' + y = t \cos(t) \]

First, solve the homogeneous equation:

\[ y'' + y = 0 \implies r^2 + 1 = 0 \implies r = \pm i \]

The complementary solution is:

\[ y = c_1 \cos(t) + c_2 \sin(t) + Y \]

Analyzing the Non-homogeneous Term

\[ f(t) = t \cos(t) \]
  • Cosine or Sine: Leads to both cosine AND sine in the particular solution \( Y \).
  • 1st-degree polynomial: The coefficient \( t \) is a first-degree polynomial.

Tentative Particular Solution

\[ Y = (At + B) \cos(t) + (Ct + D) \sin(t) \]

Observation:

  • The term \( B \cos(t) \) is copying \( c_1 \cos(t) \).
  • The term \( D \sin(t) \) is copying \( c_2 \sin(t) \).
  • But getting \( t \) copies \( At \cos(t) \).
  • So, BOTH \( At + B \) and \( Ct + D \) get an extra \( t \).

Corrected Particular Solution

\[ Y = t(At + B) \cos(t) + t(Ct + D) \sin(t) \]\[ Y' = \dots \]\[ Y'' = \dots \]
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Example: Complex Non-homogeneous Term

\[ y'' + 2y' + 5y = 3te^{-t} \cos(2t) - 2te^{-2t} \cos(t) \]

Solve the characteristic equation:

\[ r^2 + 2r + 5 = 0 \]\[ r = \frac{-2 \pm \sqrt{4 - 4(1)(5)}}{2(1)} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i \]

The complementary solution is:

\[ y = c_1 e^{-t} \cos(2t) + c_2 e^{-t} \sin(2t) + Y \]

Tentative Particular Solution

\[ Y = t \left[ (At + B)e^{-t} \cos(2t) + (Ct + D)e^{-t} \sin(2t) \right] + (Et + F)e^{-2t} \cos(t) + (Gt + H)e^{-2t} \sin(t) \]

Note on first two terms: Multiply by \( t \) for all in first two big terms because they duplicate the homogeneous solution.

Note on last two terms: Safe. No copying of the homogeneous solution.