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Given the system X' = \begin{bmatrix} -1 & 6\\ 1 & -2 \end{bmatrix}X + \begin{bmatrix} 16t\\ ...

Question:

Given the system {eq}X' = \begin{bmatrix} -1 & 6\\ 1 & -2 \end{bmatrix}X + \begin{bmatrix} 16t\\ -12 \end{bmatrix} {/eq} and the solutions for the homogeneous system {eq}X_1 = \begin{bmatrix} -2\\ 1 \end{bmatrix}e^{-4t}, X_2 = \begin{bmatrix} 3\\ 1 \end{bmatrix}e^t {/eq}, find {eq}X_p {/eq}.

How to Find the Particular Solution of a Nonhomogeneous Equations System by the Variation of Parameters Method:

The variation of parameters method is an easy technique when we have two equations in the system. This method employs the following formula to find the particular solution of the system:

{eq}\eqalign{ & {{\vec X}_p} = \phi (t)\int {{\phi ^{ - 1}}(t)} F(t)dt \cr & {\text{Where:}} \cr & \phi (t):{\text{fundamental matrix}} \cr & {\phi ^{ - 1}}(t):{\text{inverse matrix}} \cr & F(t):{\text{vector nonhomogeneous}} \cr} {/eq}

Answer and Explanation:

{eq}\eqalign{ & {\text{We use the variation of parameters method to find the particular solution of the system}}{\text{. The formula is:}} \cr & {{\vec X}_p} = \phi (t)\int {{\phi ^{ - 1}}(t)} F(t)dt \cr & {\text{We create the fundamental matrix with the two possible particular solutions.}}{\text{. The two possible solutions}} \cr & {\text{we obtain the homogeneous solutions given:}} \cr} {/eq}

{eq}\vec{x}_{h}=c_{1}\begin{pmatrix} -2\\1 \end{pmatrix}e^{-4t}+c_{2}\begin{pmatrix} 3\\1 \end{pmatrix}e^{t} {/eq}

{eq}{\text{The two possible particular solutions are:}} {/eq}

{eq}\vec{x}_{1}=\begin{pmatrix} -2e^{-4t}\\e^{-4t} \end{pmatrix} {/eq}

{eq}\vec{x}_{2}=\begin{pmatrix} 3e^{t}\\e^{t} \end{pmatrix} {/eq}

{eq}{\text{We write the Fundamental Matrix:}} {/eq}

{eq}\phi _{\left ( t \right )}=\begin{pmatrix} -2e^{-4t} &3e^{t} \\e^{-4t} &e^{t} \end{pmatrix} {/eq}

{eq}{\text{We find the Fundamental Matrix inverse}}{\text{. We remember that:}} {/eq}

{eq}A=\begin{pmatrix} a &b \\c &d \end{pmatrix}\Rightarrow A^{-1}=\frac{1}{detA}\begin{pmatrix} d &-b \\-c &a \end{pmatrix} {/eq}

{eq}{\text{Then:}} {/eq}

{eq}\phi ^{-1}=\frac{1}{det\phi}\begin{pmatrix} e^{t} &-3e^{t} \\-e^{-4t} &-2e^{-4t} \end{pmatrix} {/eq}

{eq}det\begin{pmatrix} -2e^{-4t} &3e^{t} \\e^{-4t} &e^{t} \end{pmatrix}=-2e^{-3t}-3e^{-3t}=-5e^{-3t} {/eq}

{eq}{\text{We substitute and solve:}} {/eq}

{eq}\phi ^{-1}=\frac{1}{-5e^{-3t}}\begin{pmatrix} e^{t} &-3e^{t} \\-e^{-4t} &-2e^{-4t} \end{pmatrix}=-\frac{1}{5}e^{3t}\begin{pmatrix} e^{t} &-3e^{t} \\-e^{-4t} &-2e^{-4t} \end{pmatrix}=\frac{1}{5}\begin{pmatrix} -e^{4t} &3e^{4t} \\e^{-t} &2e^{-t} \end{pmatrix} {/eq}

{eq}{\text{We replace in }}{{\vec X}_p}{\text{ formula:}} {/eq}

{eq}\vec{X}_{p}=\begin{pmatrix} -2e^{-4t} &3e^{t} \\e^{-4t} &e^{t} \end{pmatrix}\int \frac{1}{5}\begin{pmatrix} -e^{4t} &3e^{4t} \\e^{-t} &2e^{-t} \end{pmatrix}\cdot \begin{pmatrix} 16t\\-12 \end{pmatrix}dt {/eq}

{eq}\vec{X}_{p}=\frac{1}{5}\begin{pmatrix} -2e^{-4t} &3e^{t} \\e^{-4t} &e^{t} \end{pmatrix}\int \begin{pmatrix} -16te^{4t}-36e^{4t}\\16te^{-t}-24e^{-t} \end{pmatrix}dt {/eq}

{eq}\eqalign{ & {\text{We order and integrate by parts:}} \cr & {I_1} = \int {\left( { - 16t{e^{4t}} - 36{e^{4t}}} \right)} dt = - 16\int {t{e^{4t}}dt} - 36\int {{e^{4t}}dt} \cr & u = t\,\,\, \Rightarrow \,\,\,du = dt\,\,\, \Rightarrow \,\,\,\int {dv = \int {{e^{4t}}dt\,\,\, \Rightarrow \,\,\,\,v = \frac{1}{4}{e^{4t}}} } \cr & {I_1} = - 16\left( {\frac{1}{4}t{e^{4t}} - \frac{1}{4}\int {{e^{4t}}dt\,} } \right) - 36\int {{e^{4t}}dt\,} = - 16\left( {\frac{1}{4}t{e^{4t}} - \frac{1}{{16}}{e^{4t}}} \right) - \frac{{36}}{4}{e^{4t}} \cr & {I_1} = - 4t{e^{4t}} + {e^{4t}} - 9{e^{4t}} = - 4t{e^{4t}} - 8{e^{4t}} \cr & {I_2} = \int {\left( {16t{e^{ - t}} - 24{e^{ - t}}} \right)} dt = 16\int {t{e^{ - t}}dt} - 24\int {{e^{ - t}}dt} \cr & u = t\,\,\, \Rightarrow \,\,\,du = dt\,\,\, \Rightarrow \,\,\,\int {dv = \int {{e^{ - t}}dt\,\,\, \Rightarrow \,\,\,\,v = - {e^{ - t}}} } \cr & {I_2} = 16\left( { - t{e^{ - t}} + \int {{e^{ - t}}dt\,} } \right) - 24\int {{e^{ - t}}dt\,} = 16\left( { - t{e^{ - t}} - {e^{ - t}}} \right) + 24{e^{ - t}} \cr & {I_2} = - 16t{e^{ - t}} - 16{e^{ - t}} + 24{e^{ - t}} = - 16t{e^{ - t}} + 8{e^{ - t}} \cr & {\text{Therefore:}} \cr} {/eq}

{eq}\vec{X}_{p}=\frac{1}{5}\begin{pmatrix} -2e^{-4t} &3e^{t} \\e^{-4t} &e^{t} \end{pmatrix}\cdot \begin{pmatrix} -4te^{4t}-8e^{4t}\\-16te^{-t}+8e^{-t} \end{pmatrix} {/eq}

{eq}{\text{We multiply the matrix by the vector:}} {/eq}

{eq}\vec{X}_{p}=\frac{1}{5}\begin{pmatrix} 8t+16-48t+24\\-4t-8-16t+8 \end{pmatrix}=\frac{1}{5}\begin{pmatrix} -40t+40\\-20t \end{pmatrix} {/eq}

{eq}{\text{Multiplying by }}\frac{1}{5}{\text{ we obtain the particular solution:}} {/eq}

{eq}\vec{X}_{p}=\begin{pmatrix} -8t+8\\-4t \end{pmatrix} {/eq}


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System of Linear Equations: Definition & Examples

from High School Algebra II: Tutoring Solution

Chapter 8 / Lesson 8
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