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4.3 The Method of Undetermined Coefficients

\[y^{(n)} + p_1(t)y^{(n-1)} + p_2(t)y^{(n-2)} + \dots + p_n(t)y = g(t)\]

Usually used if \(p_1, p_2, p_3, \dots\) are constants AND \(g(t)\) is of the right form:

  • polynomial, \(\sin\) or \(\cos\), exponential
  • + products of these

Example

\[y^{(4)} + y''' = \sin 2t\]
\[r^4 + r^3 = 0 \quad \implies \quad r = 0, 0, 0, -1\]
\[y = c_1 + c_2 t + c_3 t^2 + c_4 e^{-t} + Y\]

Note: \(Y\) is the particular solution.

\[Y = A \sin 2t + B \cos 2t\]
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Derivatives of the Particular Solution

\[Y' = 2A \cos 2t - 2B \sin 2t\]\[Y'' = -4A \sin 2t - 4B \cos 2t\]\[Y''' = -8A \cos 2t + 8B \sin 2t\]\[Y^{(4)} = 16A \sin 2t + 16B \cos 2t\]

Substitution and Solving for Coefficients

Substituting into the original differential equation:

\[16A \sin 2t + 16B \cos 2t - 8A \cos 2t + 8B \sin 2t = \sin 2t\]

Equating coefficients:

\[16A + 8B = 1\]\[16B - 8A = 0\]\[32B - 16A = 0\]
\[40B = 1 \implies B = \frac{1}{40}, \quad A = \frac{1}{20}\]

General Solution

\[y = c_1 + c_2 t + c_3 t^2 + c_4 e^{-t} + \frac{1}{20} \sin 2t + \frac{1}{40} \cos 2t\]
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Example: Solving a Third-Order Linear Non-Homogeneous Differential Equation

1. The Differential Equation

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

2. Solving the Homogeneous Equation

First, we find the characteristic equation:

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

Factoring by grouping:

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

The roots are:

\[ r = -2, -1, 1 \]

The homogeneous solution is:

\[ y_h = C_1 e^t + C_2 e^{-t} + C_3 e^{-2t} \]

3. Finding the Particular Solution \( Y \)

The general form for the particular solution is:

\[ Y = At^2 + Bt + C + Dte^t \]

For \( t^2 \) in \( g(t) \):

\( At^2 + Bt + C \)

For \( e^t \):

\( Dte^t \) (extra \( t \) because of \( e^t \) in the homogeneous solution)

Calculating derivatives of \( Y \):

\[ Y' = 2At + B + Dte^t + De^t \]\[ Y'' = 2A + Dte^t + 2De^t \]\[ Y''' = Dte^t + 3De^t \]
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4. Substitution into the Differential Equation

Substitute \( Y, Y', Y'', Y''' \) into \( y''' + 2y'' - y' - 2y = e^t + t^2 \):

\[ (Dte^t + 3De^t) + 2(2A + Dte^t + 2De^t) - (2At + B + Dte^t + De^t) - 2(At^2 + Bt + C + Dte^t) = e^t + t^2 \]

Simplifying and grouping terms:

\[ 6De^t - 2At^2 - 2At - 2Bt + 4A - B - 2C = e^t + t^2 \]

5. Solving for Coefficients

Equating coefficients of like terms:

  • For \( e^t \): \( 6D = 1 \implies D = \frac{1}{6} \)
  • For \( t^2 \): \( -2A = 1 \implies A = -\frac{1}{2} \)
  • For \( t \): \( -2A - 2B = 0 \implies B = \frac{1}{2} \)
  • Constants: \( 4A - B - 2C = 0 \implies C = -\frac{5}{4} \)

Thus, the particular solution is:

\[ Y = -\frac{1}{2}t^2 + \frac{1}{2}t - \frac{5}{4} + \frac{1}{6}te^t \]

6. Final General Solution

General Solution:

\[ y = C_1 e^t + C_2 e^{-t} + C_3 e^{-2t} - \frac{1}{2}t^2 + \frac{1}{2}t - \frac{5}{4} + \frac{1}{6}te^t \]
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Initial Conditions and System of Equations

Initial Conditions (IC's)

\[ y(0)=0, \quad y'(0)=1, \quad y''(0)=0 \]

Applying Initial Conditions

For \( y(0)=0 \):

\[ 0 = C_1 + C_2 + C_3 - \frac{5}{4} \quad \rightarrow \quad C_1 + C_2 + C_3 = \frac{5}{4} \]

First Derivative \( y' \):

\[ y' = C_1 e^t - C_2 e^{-t} - 2 C_3 e^{-2t} - t + \frac{1}{2} + \frac{1}{6} t e^t + \frac{1}{6} e^t \] \[ 1 = C_1 - C_2 - 2 C_3 + \frac{1}{2} + \frac{1}{6} \quad \rightarrow \quad C_1 - C_2 - 2 C_3 = \frac{1}{3} \]

Second Derivative \( y'' \):

\[ y'' = C_1 e^t + C_2 e^{-t} + 4 C_3 e^{-2t} - 1 + \frac{1}{6} t e^t + \frac{1}{3} e^t \] \[ 0 = C_1 + C_2 + 4 C_3 - 1 + \frac{1}{3} \quad \rightarrow \quad C_1 + C_2 + 4 C_3 = \frac{2}{3} \]

System of Equations in Matrix Form

\[ \begin{cases} C_1 + C_2 + C_3 = \frac{5}{4} \\ C_1 - C_2 - 2 C_3 = \frac{1}{3} \\ C_1 + C_2 + 4 C_3 = \frac{2}{3} \end{cases} \Rightarrow \left[\begin{array}{ccc:c} 1 & 1 & 1 & 5/4 \\ 1 & -1 & -2 & 1/3 \\ 1 & 1 & 4 & 2/3 \end{array}\right] \]
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Solution and Method of Undetermined Coefficients

Final Solution for Constants

From Matlab:

\[ C_1 = 25/36, \quad C_2 = 3/4, \quad C_3 = -7/36 \]

The general solution is:

\[ y = \frac{25}{36} e^t + \frac{3}{4} e^{-t} - \frac{7}{36} e^{-2t} - \frac{1}{2} t^2 + \frac{1}{2} t - \frac{5}{4} + \frac{1}{6} t e^t \]

Example: Form of Particular Solution

Find the form of \( Y \) only for the equation:

\[ y''' - 3y'' + 2y' = t + e^t \]

Characteristic Equation:

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

General Solution Form:

\[ y = C_1 + C_2 e^t + C_3 e^{2t} + Y \]

Form of Particular Solution \( Y \):

\[ Y = (At + B)t + C e^t \cdot t \]

(Note: The term \( D t e^{2t} \) was crossed out in the original notes.)

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Example: Method of Undetermined Coefficients

Roots of the characteristic equation are:

\[ r = 0, 0, 0, \pm i, \pm i, -1 \]

Given the non-homogeneous term:

\[ g(t) = t^2 + 4 + e^t + \sin t + \cos 2t \]

Note: \( t^2 + 4 \) is a 2nd degree polynomial.

Find the form of the particular solution \( Y = ? \)

The complementary solution \( y_c \) is:

\[ y = C_1 + C_2 t + C_3 t^2 + C_4 \cos t + C_5 \sin t + C_6 t \cos t + C_7 t \sin t + C_8 e^{-t} + Y \]

The form of the particular solution \( Y \) is:

\[ Y = (At^2 + Bt + C)t^3 + De^t + (E \cos t + F \sin t)t^2 + G \cos 2t + H \sin 2t \]