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Q8E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 341

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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

The general solution is $y (x) = C_{1} x + C_{2} x \ln x + C_{3} x^{3} - x^{2}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 $x^{3} y^{'''} - 2 x^{2} y^{''} + 3 x y^{'} - 3 y = 0$ -------(1)

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

So, the complementary function is $y_{c} (x) = C_{1} x + C_{2} x \ln x + C_{3} x^{3}$

Step 3: Calculate Wornkians

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

$W [x, x \ln x, x^{3}] = |\begin{matrix} x & x \ln x & x^{3} \\ 1 & 1 + \ln x & 3 x^{2} \\ 0 & \frac{1}{x} & 6 x \end{matrix}| = 4 x^{2}$

$W_{1} = (- 1)^{3 - 1} W [\begin{matrix} x \ln x & x^{3} \end{matrix}] = |\begin{matrix} x \ln x & x^{3} \\ 1 + \ln x & 3 x^{2} \end{matrix}| = 2 x^{3} \ln x - x^{3} W_{2} = (- 1)^{3 - 2} W [\begin{matrix} x & x^{3} \end{matrix}] = 2 x^{2} W_{3} = (- 1)^{3 - 3} W [\begin{matrix} x & x \ln x \end{matrix}] = x$

Step 4: Calculate Vi

Evaluate.

$v_{1} (x) = \int \frac{g (x) W_{1}}{W} d x = \int \frac{\frac{1}{x} (2 x^{3} \ln x - x^{3})}{4 x^{2}} d x = \frac{1}{4} (2 x \ln x - 3 x)$

$v_{2} (x) = \int \frac{g (x) W_{2}}{W} d x = \int \frac{\frac{1}{x} (- 2 x^{3})}{4 x^{2}} d x = - \frac{x}{2} v_{3} (x) = \int \frac{g (x) W_{3}}{W} d x = - \frac{1}{4 x}$

Step 5: Particular solution

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

$y_{p} (x) = v_{1} (x) x + v_{2} (x) x \ln x + v_{3} (x) x^{3}$

The particular solution is,

$y_{p} (x) = v_{1} (x) x + v_{2} (x) x \ln x + v_{3} (x) x^{3} = \frac{1}{4} (2 x \ln x - 3 x) \cdot x + (- \frac{x}{2}) \cdot x \ln x + (- \frac{1}{4 x}) \cdot x^{3} = \frac{1}{2} x^{2} \ln x - \frac{3}{4} x^{2} - \frac{1}{2} x^{2} \ln x - \frac{1}{4} x^{2} = - x^{2}$

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

$y (x) = y_{c} (x) + y_{p} (x) = C_{1} x + C_{2} x \ln x + C_{3} x^{3} - x^{2}$

Most popular questions for Math Textbooks

Use the annihilator method to show that if $f (x)$ in (4) has the form $f (x) = a c o s β x + b s i n β x,$

then equation (4) has a particular solution of the form

(18) $y_{p} (x) = x^{s} {A c o s β x + B s i n β x}$ ,where s is chosen to be the smallest nonnegative integer such that $x^{3} c o s β x$ and $x^{3} \sin β x$ are not solutions to the corresponding homogeneous equation

Let y₁x₂= Ce^rx, where C (≠0) and r are real numbers,be a solution to a differential equation. Supposewe cannot determine r exactly but can only approximateit by $r$ . Let (x) =Ce^rxand consider the error

$| y (x) - \tilde{y} (x) |$

(a) If r and $\tilde{r}$ are positive, r ≠ , show that the errorgrows exponentially large as x approaches + ∞.

(b) If r and $r$ are negative, r≠ , show that the errorgoes to zero exponentially as x approaches + ∞.

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

${x^{2}, x^{2} - 1, 5}$ on $(- \infty, \infty)$

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

$y^{'''} - 3 y^{''} + 3 y^{'} - y = e^{x}$

Higher-Order Cauchy–Euler Equations. A differential equation that can be expressed in the form

$a_{n} x^{n} y^{(n)} (x) + a_{n - 1} x^{n - 1} y^{(n - 1)} (x) + . .. + a_{0} y (x) = 0$

where $a_{n}, a_{n - 1}, . ..., a_{0}$ are constants, is called a homogeneous Cauchy–Euler equation. (The second-order case is discussed in Section 4.7.) Use the substitution $y = x^{y}$ to help determine a fundamental solution set for the following Cauchy–Euler equations:

(a) $x^{3} y''' + x^{2} y'' - 2 x y' + 2 y = 0, x > 0$

(b) $x^{4} y^{(4)} + 6 x^{3} y''' + 2 x^{2} y'' - 4 x y + 4 y = 0, x > 0$

(c) $x^{3} y''' - 2 x^{2} y'' + 13 x y' - 13 y = 0, x > 0$

[Hint: $x^{α + β i} = e^{(α + β i) l n x} = x^{a} {c o s (β l n x) + i s i n (β l n x)}$ ]

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