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7.9 Nonhomogeneous Systems

\[ \vec{x}' = A\vec{x} + \vec{g}(t) \]

methods we will look at: undetermined coeff
variation of parameters

Example

\[ \vec{x}' = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \vec{x} + \begin{bmatrix} 3e^{2t} \\ 2t \end{bmatrix} \quad \text{(undetermined coeff)} \]

First solve the homogeneous system \( \vec{x}' = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \vec{x} \)

General solution: \( \vec{x} = \vec{x}_c + \vec{x}_p \)

Where \( \vec{x}_c \) is the complementary (homogeneous) solution and \( \vec{x}_p \) is the particular solution (due to \( \vec{g}(t) \)).

\[ \vec{x}_c = c_1 e^t \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

\( \vec{x}_p \) resembles \( \vec{g}(t) \) in form

\[ \vec{g}(t) = \begin{bmatrix} 3e^{2t} \\ 2t \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} e^{2t} + \begin{bmatrix} 0 \\ 2 \end{bmatrix} t \]

Where \( \begin{bmatrix} 3 \\ 0 \end{bmatrix} e^{2t} \) is the exponential part and \( \begin{bmatrix} 0 \\ 2 \end{bmatrix} t \) is the linear polynomial part.

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\[ \vec{x}_p = \vec{a} e^{2t} + \vec{b} t + \vec{c} \]

Where \( \vec{a} \) is the undetermined vector coefficient and \( \vec{b} t + \vec{c} \) is due to the \( \begin{bmatrix} 0 \\ 2 \end{bmatrix} t \) term.

Find \( \vec{a}, \vec{b}, \vec{c} \) by substituting \( \vec{x}_p \) into \( \vec{x}' = A\vec{x} + \vec{g} \)

Linear equation so we can consider the parts of \( \vec{x}_p \) separately.

Solve \( \vec{x}' = A\vec{x} + \vec{g} \) with parts due to \( e^{2t} \) only

\[ \vec{x}_p = \vec{a} e^{2t} \] \[ \vec{x}_p' = 2 \vec{a} e^{2t} \]

\( \vec{x}' = A\vec{x} + \vec{g} \) (\( e^{2t} \) part only):

\[ 2 \vec{a} e^{2t} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \vec{a} e^{2t} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} e^{2t} \quad \text{where } \vec{a} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} \] \[ \begin{bmatrix} 2a_1 \\ 2a_2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \]

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Solving for Particular Solution Coefficients

\[ \begin{bmatrix} 2a_1 \\ 2a_2 \end{bmatrix} = \begin{bmatrix} 2a_1 + a_2 + 3 \\ a_1 + 2a_2 \end{bmatrix} \quad \begin{matrix} 2a_1 = 2a_1 + a_2 + 3 & \rightarrow & a_2 = -3 \\ 2a_2 = a_1 + 2a_2 & \rightarrow & a_1 = 0 \end{matrix} \]

Repeat for the \( \begin{bmatrix} 0 \\ 2 \end{bmatrix} t \) part of \( \vec{g} \)

\[ \begin{aligned} \vec{x}_p &= \vec{b}t + \vec{c} \\ \vec{x}_p' &= \vec{b} \end{aligned} \]

Sub into \( \vec{x}' = A\vec{x} + \vec{g} \) (only \( t \) part):

\[ \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} t + \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 2 \end{bmatrix} t \]

Equating Coefficients

\[ \text{terms w/ } t: \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 2b_1 + b_2 \\ b_1 + 2b_2 + 2 \end{bmatrix} \implies \begin{matrix} b_1 = \frac{2}{3} \\ b_2 = -\frac{4}{3} \end{matrix} \]

\[ \text{terms w/o } t: \begin{bmatrix} b_1 \\ b_2 \end{bmatrix} = \begin{bmatrix} 2c_1 + c_2 \\ c_1 + 2c_2 \end{bmatrix} \implies \begin{matrix} c_1 = \frac{8}{9} \\ c_2 = -\frac{10}{9} \end{matrix} \]

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General Solution and Duplication

General Solution

\[ \vec{x} = \underbrace{c_1 e^t \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}}_{\vec{x}_c} + \underbrace{\begin{bmatrix} 0 \\ -3 \end{bmatrix} e^{2t} + \begin{bmatrix} 2/3 \\ -4/3 \end{bmatrix} t + \begin{bmatrix} 8/9 \\ -10/9 \end{bmatrix}}_{\vec{x}_p} \]

Where the vectors in the particular solution correspond to \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) respectively.

What about duplication?

\[ \vec{x}' = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \vec{x} + \begin{bmatrix} 3 \\ 3 \end{bmatrix} e^{3t} \]

\[ \vec{x}_c = c_1 e^t \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

Note: \( \vec{g} \) copies the second term of \( \vec{x}_c \).

Initial guess for \( \vec{x}_p \): \( \vec{x}_p = \vec{a} e^{3t} \)

Fix for \( \vec{x}_p \): \( \vec{x}_p = \vec{a} t e^{3t} + \vec{b} e^{3t} \)

Just like when eigenvalues are repeated without enough eigenvectors.

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Method of Undetermined Coefficients: Repeated Roots

potentially very tedious if copied multiple times

\[ \vec{x}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \vec{x} + \begin{bmatrix} 0 \\ 2e^t \end{bmatrix} \]

\[ \begin{aligned} \vec{x}_c &= c_1 e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^t \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \\ &= c_1 e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 t e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^t \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{aligned} \]

Initial \( \vec{x}_p \): \( \vec{x}_p = \vec{a} e^t \) copying parts of \( \vec{x}_c \)

First fix: \( \vec{x}_p = \vec{a} t e^t + \vec{b} e^t \) still copying

Fix again: \( \vec{x}_p = \vec{a} t^2 e^t + \vec{b} t e^t \)

ok if \( \vec{b} \) is linearly independent from \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) (eigenvector)

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Variation of Parameters

\[ \vec{x}' = A \vec{x} + \vec{g}(t) \]

\[ \vec{x}_c = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 \]

Complementary (\( \vec{x}' = A \vec{x} \) only)

\[ = \Psi(t) \vec{c} \]

Where \( \vec{c} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \)

Fundamental matrix

\( e^{\lambda t} \vec{v} \) as columns

Variation of Parameters: \( \vec{x} = \Psi(t) \vec{u}(t) \)

Where \( \vec{u}(t) = \begin{bmatrix} u_1(t) \\ u_2(t) \end{bmatrix} \)

How to find \( \vec{u} \)?

Sub \( \vec{x} = \Psi(t) \vec{u} \) into \( \vec{x}' = A \vec{x} + \vec{g} \)

\[ \vec{x}' = \Psi \vec{u}' + \Psi' \vec{u} \]

\[ \Psi \vec{u}' + \Psi' \vec{u} = A \Psi \vec{u} + \vec{g} \]

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Variation of Parameters for Systems

Terms with \( \vec{u} \): \( \Psi' = A\Psi \) → reflecting \( \vec{x}' = A\vec{x} \)

Terms w/o \( \vec{u} \): \( \Psi \vec{u}' = \vec{g} \)

Scalar case:

\( u_1' y_1 + u_2' y_2 = 0 \)
\( u_1' y_1' + u_2' y_2' = g \)

Example

\[ \vec{x}' = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \vec{x} + \begin{bmatrix} 3 \\ 3 \end{bmatrix} e^{3t} \]

\[ \vec{x}_c = c_1 e^t \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

\[ \Psi = \begin{bmatrix} e^t & e^{3t} \\ -e^t & e^{3t} \end{bmatrix} \]

Solve \( \Psi \vec{u}' = \vec{g} \)

\[ \left[ \begin{array}{cc|c} e^t & e^{3t} & 3e^{3t} \\ -e^t & e^{3t} & 3e^{3t} \end{array} \right] \]

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\[ \xrightarrow{R_1 + R_2} \left[ \begin{array}{cc|c} e^t & e^{3t} & 3e^{3t} \\ 0 & 2e^{3t} & 6e^{3t} \end{array} \right] \]

Row 2: \( 2e^{3t} u_2' = 6e^{3t} \)

\( u_2' = 3 \quad \implies \quad u_2 = 3t + c_2 \)

Row 1: \( e^t u_1' + e^{3t} u_2' = 3e^{3t} \)

\( e^t u_1' + e^{3t} \cdot 3 = 3e^{3t} \)

\( u_1' = 0 \quad \implies \quad u_1 = c_1 \)

Solution

\[ \Psi \vec{u} = \begin{bmatrix} e^t & e^{3t} \\ -e^t & e^{3t} \end{bmatrix} \begin{bmatrix} c_1 \\ 3t + c_2 \end{bmatrix} \]

\[ = c_1 \begin{bmatrix} e^t \\ -e^t \end{bmatrix} + c_2 \begin{bmatrix} e^{3t} \\ e^{3t} \end{bmatrix} + 3t \begin{bmatrix} e^{3t} \\ e^{3t} \end{bmatrix} \]