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Q31E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 332

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

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

${x^{1}, x^{2} {c o s (3 l n x) + i s i n (3 l n x)}, x^{2} {c o s (3 l n x) - i s i n (3 l n x)}$ is the fundamental solution set.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Solving for (a):

(a)Given differential equation is,

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

Let,

$\begin{array}{l} y = x^{r} \\ y' = r x^{r - 1} \\ y'' = r (r - 1) x^{r - 2} \\ y''' = r (r - 1) (r - 2) x^{r - 3} \end{array}$

Substitution in equation (1) we get,

$\begin{array}{l} (r x^{r - 1}) + 2 x^{r} = 0 \\ r (r - 1) (r - 2) x^{r - 3 + 3} + r (r - 1) x^{r - 2 + 2} - 2 r x^{r - 1 + 1} + 2 x^{r} = 0 \\ r (r - 1) (r - 2) x^{r} + r (r - 1) x^{r} - 2 r x^{r} + 2 x^{r} = 0 \\ (r (r - 1) (r - 2) + r (r - 1) - 2 r + 2) x^{r} = 0 \end{array}$

Since $x > 0, x^{r} > 0$

$\begin{array}{l} (r (r - 1) (r - 2) + r (r - 1) - 2 r + 2) = 0 \\ r (r - 1) (r - 2) + r (r - 1) - 2 (r - 1) = 0 \\ (r - 1) (r (r - 2) + r - 2) = 0 \\ (r - 1) (r (r - 2) + (r - 2)) = 0 \\ (r - 1) (r - 2) (r + 1) = 0 \\ \Rightarrow r = 1, r = 2, r = - 1 \end{array}$

Therefore,

$y_{1} = x^{1}, y_{2} = x^{2}, y_{3} = x^{- 1}$

Are all solutions of equation (1)

${x^{1}, x^{2}, x^{- 1}}$ is the fundamental solution set.

Step 2: Solution for (b):

(b)Given differential equation is,

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

Let,

$\begin{array}{l} y = x^{r} \\ y' = r x^{r - 1} \\ y'' = r (r - 1) x^{r - 2} \\ y''' = r (r - 1) (r - 2) x^{r - 3} \\ y'''' = r (r - 1) (r - 2) (r - 3) x^{r - 4} \end{array}$

Substituting in equation (2), we get

$\begin{array}{l} x^{4} (r (r - 1) (r - 2) (r - 3) x^{r - 4}) + 6 x^{3} (r (r - 1) (r - 2) x^{r - 3}) + 2 x^{2} (r (r - 1) x^{r - 2}) - 4 x (r x^{r - 1}) + 4 (x^{r}) = 0 \\ r (r - 1) (r - 2) (r - 3) x^{r - 4 + 4} + 6 r (r - 1) (r - 3) x^{r - 3 + 3} + 2 r (r - 1) x^{r - 2 + 2} - 4 r x^{r - 1 + 1} + 4 x^{r} = 0 \\ r (r - 1) (r - 2) (r - 3) x^{r} + 6 r (r - 1) (r - 3) x^{r} + 2 r (r - 1) x^{r} - 4 r x^{r} + 4 x^{r} = 0 \\ (r (r - 1) (r - 2) (r - 3) + 6 r (r - 1) (r - 3) + 2 r (r - 1) - 4 r + 4) x^{r} = 0 \end{array}$

Since $x > 0, x^{r} > 0$

$\begin{array}{l} \Rightarrow r (r - 1) (r - 2) (r - 3) + 6 r (r - 1) (r - 3) + 2 r (r - 1) - 4 r + 4 = 0 \\ r (r - 1) (r - 2) (r - 3) + 6 r (r - 1) (r - 3) + 2 r (r - 1) - 4 (r - 1) = 0 \\ (r - 1) (r (r - 2) (r - 3) + 6 r (r - 2) - 2 r - 4) = 0 \\ (r - 1) (r (r - 2) (r - 3) + 6 r (r - 2) + 2 (r - 2)) = 0 \\ (r - 1) ((r - 2) (r (r - 3) + 6 r + 2)) = 0 \\ (r - 1) (r - 2) (r (r - 3) + 6 r + 2) = 0 \\ (r - 1) (r - 2) (r^{2} - 3 r + 6 r + 2) = 0 \\ (r - 1) (r - 2) (r^{2} - 3 r + 2) = 0 \\ (r - 1) (r - 2) (r + 1) (r + 2) = 0 \\ \Rightarrow r = 1, r = 2, r = - 1, r = - 2 \end{array}$

Therefore,

$y_{1} = x^{1}, y_{2} = x^{2}, y_{3} = x^{- 1}, y_{4} = x^{- 2}$

Are all solution of equation (2)

Step 3: Solving further:

$\begin{array}{l} y_{3} = x^{2 - 3 i} \\ = e^{(2 - 3 i) l n x} \\ = x^{2} {c o s (- 3 l n x) + i s i n (- 3 l n x)} \\ = x^{2} {c o s (3 l n x) - i s i n (3 l n x)} \end{array}$

Hence,

${x^{1}, x^{2} {c o s (3 l n x) + i s i n (3 l n x)}, x^{2} {c o s (3 l n x) - i s i n (3 l n x)}$ is the fundamental solution set.

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