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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 # Find a general solution to the Cauchy-Euler equation ${{\mathbit{x}}}^{{\mathbf{3}}}{{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{{\mathbit{x}}}^{{\mathbf{2}}}{{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{3}}{\mathbit{x}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{-}}{\mathbf{3}}{\mathbit{y}}{\mathbf{=}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{,}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbit{x}}{\mathbf{>}}{\mathbf{0}}{\mathbf{,}}$given that $\left\{x,\mathrm{xlnx},{x}^{3}\right\}$ is a fundamental solution set for the corresponding homogeneous equation

The general solution is $y\left(x\right)={C}_{1}x+{C}_{2}x\mathrm{ln}x+{C}_{3}{x}^{3}-{x}^{2}$

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

.

It is given that ${x}^{3}{y}^{\text{'}\text{'}\text{'}}-2{x}^{2}{y}^{\text{'}\text{'}}+3x{y}^{\text{'}}-3y=0$ -------(1)

The fundamental solution set is $\left\{x,x\mathrm{ln}x,{x}^{3}\right\}$

So, the complementary function is ${y}_{c}\left(x\right)={C}_{1}x+{C}_{2}x\mathrm{ln}x+{C}_{3}{x}^{3}$

## Step 3: Calculate Wornkians

Find $W,{W}_{k},k=1,2,3$as follows:

$W\left[x,x\mathrm{ln}x,{x}^{3}\right]=\left|\begin{array}{ccc}x& x\mathrm{ln}x& {x}^{3}\\ 1& 1+\mathrm{ln}x& 3{x}^{2}\\ 0& \frac{1}{x}& 6x\end{array}\right|\phantom{\rule{0ex}{0ex}}=4{x}^{2}\phantom{\rule{0ex}{0ex}}$

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

## Step 4: Calculate Vi

Evaluate.

${v}_{1}\left(x\right)=\int \frac{g\left(x\right){W}_{1}}{W}dx\phantom{\rule{0ex}{0ex}}=\int \frac{\frac{1}{x}\left(2{x}^{3}\mathrm{ln}x-{x}^{3}\right)}{4{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(2x\mathrm{ln}x-3x\right)\phantom{\rule{0ex}{0ex}}$

${v}_{2}\left(x\right)=\int \frac{g\left(x\right){W}_{2}}{W}dx\phantom{\rule{0ex}{0ex}}=\int \frac{\frac{1}{x}\left(-2{x}^{3}\right)}{4{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=-\frac{x}{2}\phantom{\rule{0ex}{0ex}}{v}_{3}\left(x\right)=\int \frac{g\left(x\right){W}_{3}}{W}dx\phantom{\rule{0ex}{0ex}}=-\frac{1}{4x}\phantom{\rule{0ex}{0ex}}$

## Step 5: Particular solution

Since $\left\{x,x\mathrm{ln}x,{x}^{3}\right\}$ is a fundamental solution set, so a particular solution of the form,

${y}_{p}\left(x\right)={v}_{1}\left(x\right)x+{v}_{2}\left(x\right)x\mathrm{ln}x+{v}_{3}\left(x\right){x}^{3}$

The particular solution is,

${y}_{p}\left(x\right)={v}_{1}\left(x\right)x+{v}_{2}\left(x\right)x\mathrm{ln}x+{v}_{3}\left(x\right){x}^{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(2x\mathrm{ln}x-3x\right)·x+\left(-\frac{x}{2}\right)·x\mathrm{ln}x+\left(-\frac{1}{4x}\right)·{x}^{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{x}^{2}\mathrm{ln}x-\frac{3}{4}{x}^{2}-\frac{1}{2}{x}^{2}\mathrm{ln}x-\frac{1}{4}{x}^{2}\phantom{\rule{0ex}{0ex}}=-{x}^{2}$

Thus, the general solution of the equation (1) is,

$y\left(x\right)={y}_{c}\left(x\right)+{y}_{p}\left(x\right)\phantom{\rule{0ex}{0ex}}={C}_{1}x+{C}_{2}x\mathrm{ln}x+{C}_{3}{x}^{3}-{x}^{2}\phantom{\rule{0ex}{0ex}}$ ### Want to see more solutions like these? 