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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 # In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation. ${{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{3}}{{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbit{y}}{\mathbf{=}}{{\mathbit{e}}}^{\mathbf{2}\mathbf{x}}$

The particular solution is ${y}_{p}\left(x\right)=-\frac{1}{27}{e}^{2x}-\frac{2}{9}\left(\frac{3}{2}{x}^{2}+x\right)·{e}^{2x}-\frac{1}{3}{x}^{2}·{e}^{2x}$

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

The given equation is: ${y}^{\text{'}\text{'}\text{'}}-3{y}^{\text{'}\text{'}}+4y={e}^{2x}$

The auxiliary equation is ${m}^{3}-3{m}^{2}+4=0$

Solving for $m$ we get value:

$m=-1,2,2$

The complimentary solution is $CF={c}_{1}{e}^{-x}+{c}_{2}{e}^{2x}+{c}_{3}x{e}^{2x}$

## Step 3: Calculate Wornkians

Compare with $CF={c}_{1}{y}_{1}\left(x\right)+{c}_{2}{y}_{2}\left(x\right)+{c}_{3}{y}_{3}\left(x\right)$ .

${y}_{1}\left(x\right)={e}^{-x},{y}_{2}\left(x\right)={e}^{2x}\text{and}{y}_{3}\left(x\right)=x{e}^{2x}$

Find four Wronkians of determinant.

$W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)=\left|\begin{array}{c}{y}_{1}\\ y{\text{'}}_{1}\\ y{\text{'}\text{'}}_{1}\end{array}\begin{array}{c}{y}_{2}\\ y{\text{'}}_{2}\\ y{\text{'}\text{'}}_{2}\end{array}\begin{array}{c}{y}_{3}\\ y{\text{'}}_{3}\\ y{\text{'}\text{'}}_{3}\end{array}\right|\phantom{\rule{0ex}{0ex}}W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)=\left|\begin{array}{c}{}_{{e}^{-x}}\\ {-}_{{e}^{-x}}\\ {}_{{e}^{-x}}\end{array}\begin{array}{c}{}_{{e}^{2x}}\\ 2{e}^{2x}\\ 4{e}^{2x}\end{array}\begin{array}{c}x{e}^{2x}\\ 2\left(x+1\right){e}^{2x}\\ 4\left(x+1\right){e}^{2x}\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Take common factor out from each column.

$W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)={e}^{-x}·{e}^{2x}·{e}^{2x}\left|\begin{array}{c}1\\ -1\\ 1\end{array}\begin{array}{c}1\\ 2\\ 4\end{array}\begin{array}{c}x\\ \left(2x+1\right)\\ 4\left(x+1\right)\end{array}\right|$

Solving we get:

${W}_{1}\left(x\right)=\left(-1{\right)}^{3-1}W\left[{y}_{2},{y}_{3}\right]\left(x\right)={e}^{4x}$

${W}_{2}\left(x\right)=-{e}^{x}\left(3x+1\right)\phantom{\rule{0ex}{0ex}}{W}_{3}\left(x\right)=3{e}^{x}$

## Step 4: Particular solution

The particular solution is given by ${y}_{p}\left(x\right)={y}_{1}\left(x\right)·{v}_{1}\left(x\right)+{y}_{2}\left(x\right)·{v}_{2}\left(x\right)+{y}_{3}\left(x\right)·{v}_{3}\left(x\right)$

Here,

${v}_{1}\left(x\right)=-\int \frac{g\left(x\right){W}_{1}\left(x\right)}{W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)}dx\phantom{\rule{0ex}{0ex}}{v}_{1}\left(x\right)=-\int \frac{{e}^{2x}·{e}^{4x}}{9{e}^{3x}}dx\phantom{\rule{0ex}{0ex}}{v}_{1}\left(x\right)=-\frac{1}{27}{e}^{3x}\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=\int \frac{g\left(x\right){W}_{2}\left(x\right)}{W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)}dx\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=\int \frac{{e}^{2x}·\left\{-{e}^{x}\left(3x+1\right)\right\}}{9{e}^{3x}}dx\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=-\frac{1}{9}\left(\frac{3}{2}{x}^{2}+x\right)$

${v}_{1}\left(x\right)=-\int \frac{g\left(x\right){W}_{1}\left(x\right)}{W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)}dx\phantom{\rule{0ex}{0ex}}{v}_{1}\left(x\right)=-\int \frac{{e}^{2x}·{e}^{4x}}{9{e}^{3x}}dx\phantom{\rule{0ex}{0ex}}{v}_{1}\left(x\right)=-\frac{1}{27}{e}^{3x}\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=\int \frac{g\left(x\right){W}_{2}\left(x\right)}{W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)}dx\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=\int \frac{{e}^{2x}·\left\{-{e}^{x}\left(3x+1\right)\right\}}{9{e}^{3x}}dx\phantom{\rule{0ex}{0ex}}{v}_{2}\left(x\right)=-\frac{1}{9}\left(\frac{3}{2}{x}^{2}+x\right)$

${v}_{3}\left(x\right)=-\int \frac{g\left(x\right){W}_{3}\left(x\right)}{W\left[{y}_{1},{y}_{2},{y}_{3}\right]\left(x\right)}dx\phantom{\rule{0ex}{0ex}}{v}_{3}\left(x\right)=-\frac{1}{3}x$

Thus, the particular solution is given by:

${y}_{p}\left(x\right)={v}_{1}\left(x\right)·{y}_{1}\left(x\right)+{v}_{2}\left(x\right)·{y}_{2}\left(x\right)+{v}_{3}\left(x\right)·{y}_{3}\left(x\right)\phantom{\rule{0ex}{0ex}}{y}_{p}\left(x\right)=-\frac{1}{27}{e}^{2x}-\frac{2}{9}\left(\frac{3}{2}{x}^{2}+x\right)·{e}^{2x}-\frac{1}{3}{x}^{2}·{e}^{2x}$ ### Want to see more solutions like these? 