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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.${{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{\mathbit{t}}{\mathbit{a}}{\mathbit{n}}{\mathbit{x}}$

The particular solution is${y}_{p}=\mathrm{ln}|\mathrm{sec}x|-\mathrm{sin}x\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|$

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:${y}^{\text{'}\text{'}\text{'}}+{y}^{\text{'}}=\mathrm{tan}x$

The auxiliary equation is ${D}^{3}+D=0$

So, $D=±i,0$

So $\left\{1,\mathrm{cos}x,\mathrm{sin}x\right\}$fundamental set.

## Step 3: Calculate Wornkians

The value of wronkians is:

$W\left[\begin{array}{ccc}1& \mathrm{cos}x& sinx\end{array}\right]=\left|\begin{array}{ccc}1& \mathrm{cos}x& \mathrm{sin}x\\ 0& -\mathrm{sin}x& \mathrm{cos}x\\ 0& -\mathrm{cos}x& -\mathrm{sin}x\end{array}\right|\phantom{\rule{0ex}{0ex}}=1$

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

## Step 4: For particular solution

The particular solution is given by:

${y}_{p}=1\int \frac{1\left(\mathrm{tan}x\right)}{1}dx+\mathrm{cos}x\int \frac{-\mathrm{cos}x\left(\mathrm{tan}x\right)}{1}dx+\mathrm{sin}x\int \frac{-\mathrm{sin}x·\mathrm{tan}x}{1}dx\phantom{\rule{0ex}{0ex}}{y}_{p}=\mathrm{ln}|\mathrm{sec}x|+{\mathrm{cos}}^{2}x+\mathrm{sin}x{I}_{1}\phantom{\rule{0ex}{0ex}}{I}_{1}=\int -\mathrm{sin}x·\mathrm{tan}xdx\phantom{\rule{0ex}{0ex}}=\mathrm{sin}x-\mathrm{ln}\left(\mathrm{sec}x+\mathrm{tan}x\right)\phantom{\rule{0ex}{0ex}}$

${y}_{p}=\mathrm{ln}|\mathrm{sec}x|+{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{2}x\mathrm{ln}\left(\mathrm{sec}x+\mathrm{tan}x\right)$

Since 1 is in fundamental set solution so ${y}_{p}=\mathrm{ln}|\mathrm{sec}x|-\mathrm{sin}x\mathrm{ln}|\mathrm{sec}x+\mathrm{tan}x|$