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Expert-verified Found in: Page 341 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Given that$\left\{{e}^{x},{e}^{-x},{e}^{2x}\right\}$ is a fundamental solution set for the homogeneous equation corresponding to the equation ${{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{-}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{+}}{\mathbf{2}}{\mathbit{y}}{\mathbf{=}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{,}}$determine a formula involving integrals for a particular solution.

The particular solution is ${y}_{p}\left(x\right)=-\frac{{e}^{x}}{2}\int {e}^{-x}g\left(x\right)dx+\frac{{e}^{-x}}{6}\int {e}^{x}g\left(x\right)dx+\frac{{e}^{2x}}{3}\int {e}^{x}g\left(x\right)dx$

See the step by step solution

## Step 1: Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

## Step 2: Find complementary solution

Consider the differential equation ${y}^{\text{'}\text{'}\text{'}}-2{y}^{\text{'}\text{'}}-{y}^{\text{'}}+2y=g\left(x\right)$

Consider the fundamental solution of the equation as,$\left\{{e}^{x},{e}^{-x},{e}^{2x}\right\}$

Therefore the complementary solutions of the equations is ${y}_{c}={c}_{1}{e}^{x}+{c}_{2}{e}^{-x}+{c}_{3}x{e}^{2x}$

## Step 3: Wronkians

Here we have ${y}_{1}\left(x\right)={e}^{x},{y}_{2}\left(x\right)={e}^{-x},{y}_{3}\left(x\right)={e}^{2x}$

Calculates the corresponding Wronskian is,

Apply row operation, and then Wronskian is,

$W\left[{y}_{1},{y}_{2},{y}_{3}\right]=\left|\begin{array}{ccc}{e}^{x}& {e}^{-x}& {e}^{2x}\\ {e}^{x}& -{e}^{x}& 2{e}^{x}\\ {e}^{x}& {e}^{-x}& 4{e}^{2x}\end{array}\right|\phantom{\rule{0ex}{0ex}}W\left[{y}_{1},{y}_{2},{y}_{3}\right]=\left|\begin{array}{ccc}{e}^{x}& {e}^{-x}& {e}^{2x}\\ {e}^{x}& -{e}^{x}& 2{e}^{x}\\ 0& 0& 3{e}^{2x}\end{array}\right|;{R}_{3}^{\text{'}}={R}_{3}-{R}_{1}\phantom{\rule{0ex}{0ex}}=-6{e}^{2x}$

${W}_{1}\left(x\right)=\left(-1{\right)}^{3-1}\left|\begin{array}{cc}{e}^{-x}& {e}^{2x}\\ -{e}^{-x}& 2{e}^{2x}\end{array}\right|\phantom{\rule{0ex}{0ex}}=3{e}^{x}\phantom{\rule{0ex}{0ex}}{W}_{2}\left(x\right)=\left(-1{\right)}^{3-2}\left|\begin{array}{cc}{e}^{x}& {e}^{2x}\\ {e}^{x}& 2{e}^{2x}\end{array}\right|\phantom{\rule{0ex}{0ex}}=-{e}^{3x}\phantom{\rule{0ex}{0ex}}{W}_{3}\left(x\right)=\left(-1{\right)}^{3-3}\left|\begin{array}{cc}{e}^{x}& {e}^{-x}\\ {e}^{x}& -{e}^{-x}\end{array}\right|\phantom{\rule{0ex}{0ex}}=-2\phantom{\rule{0ex}{0ex}}$

## Step 4: Calculate V1

We know that ${v}_{k}\left(x\right)=\int \frac{g\left(x\right){W}_{k}\left(x\right)}{W\left[{y}_{1},{y}_{2},{y}_{3}\right]}dx$

Hence,

${v}_{1}\left(x\right)=\int \frac{g\left(x\right)3{e}^{x}}{-6{e}^{2x}}dx\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}\int g\left(x\right){e}^{-x}dx\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=-\int \frac{g\left(x\right){e}^{3x}}{-6{e}^{2x}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\int g\left(x\right){e}^{-2x}dx\phantom{\rule{0ex}{0ex}}{v}_{3}\left(x\right)=\int \frac{\left(-2\right)g\left(x\right)}{-6{e}^{2x}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\int g\left(x\right){e}^{-2x}dx\phantom{\rule{0ex}{0ex}}$

Therefore the particular solution is involving integral is:

${y}_{p}\left(x\right)=-\frac{{e}^{x}}{2}\int {e}^{-x}g\left(x\right)dx+\frac{{e}^{-x}}{6}\int {e}^{x}g\left(x\right)dx+\frac{{e}^{2x}}{3}\int {e}^{x}g\left(x\right)dx$ ### Want to see more solutions like these? 