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### 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.${{\mathbit{z}}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{3}}{{\mathbit{z}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{4}}{\mathbit{z}}{\mathbf{=}}{{\mathbit{e}}}^{\mathbf{2}\mathbf{x}}$

The particular solution is ${Z}_{p}=\frac{1}{16}{e}^{2x}$

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:${z}^{\text{'}\text{'}\text{'}}+3{z}^{\text{'}\text{'}}-4z={e}^{2x}$

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

Simplifying we get

${m}^{3}+3{m}^{2}-4=0\phantom{\rule{0ex}{0ex}}\left(m-1\right)\left(m+2{\right)}^{2}=0\phantom{\rule{0ex}{0ex}}m=1,-2,-2$

Therefore, the complimentary solution is given by ${Z}_{c}={c}_{1}{e}^{x}+{c}_{2}{e}^{-2x}+{c}_{3}x{e}^{-2x}$

## Step 3: Calculate Wornkians

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

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

$W\left(x\right)=\left|\begin{array}{ccc}{e}^{x}& {e}^{-2x}& x{e}^{-2x}\\ {\left({e}^{x}\right)}^{0}& {\left({e}^{2x}\right)}^{0}& \left(x{e}^{-2x}\right)\\ {\left({e}^{x}\right)}^{*}& \left({e}^{-2x}\right)*& \left(x{e}^{-2x}\right)*\end{array}\right|\phantom{\rule{0ex}{0ex}}={e}^{x}\left({e}^{-4x}\left(-8x+8+8x-4\right)\right)-{e}^{-2x}\left({e}^{-x}\left(4x-4+2x-1\right)\right)+x{e}^{-2x}\left(4{e}^{-x}+2{e}^{-x}\right)\phantom{\rule{0ex}{0ex}}=4{e}^{-3x}-6x{e}^{-3x}+5{e}^{-3x}+6x{e}^{-3x}\phantom{\rule{0ex}{0ex}}=9{e}^{-3x}\phantom{\rule{0ex}{0ex}}$

## Step 4: Calculate Wornkians

${W}_{1}=\left(-1{\right)}^{3-1}W\left({e}^{-2x},x{e}^{-2x}\right)\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{cc}{e}^{-2x}& x{e}^{-2x}\\ -2{e}^{-2x}& {e}^{-2x}\left(1-2x\right)\end{array}\right|\phantom{\rule{0ex}{0ex}}={e}^{-4x}\phantom{\rule{0ex}{0ex}}{W}_{2}=\left(-1{\right)}^{3-2}W\left({e}^{x},x{e}^{-2x}\right)\phantom{\rule{0ex}{0ex}}=-\left|\begin{array}{cc}{e}^{x}& x{e}^{-2x}\\ {e}^{x}& {e}^{-2x}\left(1-2x\right)\end{array}\right|\phantom{\rule{0ex}{0ex}}=-{e}^{-x}\left(1-3x\right)\phantom{\rule{0ex}{0ex}}$

And the value${w}_{3}$is,

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

## 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,

${V}_{1}=\int \frac{{W}_{1}\left(x\right)f\left(x\right)}{W\left(x\right)}dx\phantom{\rule{0ex}{0ex}}=\frac{{e}^{x}}{9}\phantom{\rule{0ex}{0ex}}{V}_{2}=\int \frac{{W}_{2}\left(x\right)f\left(x\right)}{W\left(x\right)}dx\phantom{\rule{0ex}{0ex}}=\int \frac{-{e}^{-x}\left(1-3x\right){e}^{2x}}{9{e}^{-3x}}dx\phantom{\rule{0ex}{0ex}}=-\frac{7{e}^{4x}}{144}+\frac{x{e}^{4x}}{12}\phantom{\rule{0ex}{0ex}}$

And

${V}_{3}=\int \frac{{W}_{3}\left(x\right)f\left(x\right)}{W\left(x\right)}dx\phantom{\rule{0ex}{0ex}}=-\frac{{e}^{4x}}{12}\phantom{\rule{0ex}{0ex}}$

Therefore, the particular integral is given by,

${Z}_{p}={V}_{1}{e}^{x}+{V}_{2}{e}^{-2x}+{V}_{3}x{e}^{-2x}\phantom{\rule{0ex}{0ex}}={e}^{x}\frac{{e}^{x}}{9}+\left(\frac{{e}^{4x}}{144}+\frac{x{e}^{4x}}{12}\right){e}^{-2x}+\left(-\frac{{e}^{4x}}{12}\right)x{e}^{-2x}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{2x}}{16}\phantom{\rule{0ex}{0ex}}$

Therefore the particular solution is ${Z}_{p}=\frac{1}{16}{e}^{2x}$