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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 6: Q7E (page 326)

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.
${e^{3 x}, 鈥� e^{5 x}, 鈥� e^{- x}}$ on $(- 鈭�, 鈥� 鈭�)$

Short Answer

Expert verified

Thus, $\{e^{3 x}, 鈥� e^{5 x}, 鈥� e^{- x}\}$ are linearly independent on $(- 鈭�, 鈥� 鈭�)$ .

Step by step solution

01

Step 1:Using the concept of Wronskian

The given function is $\{e^{3 x}, 鈥� e^{5 x}, 鈥� e^{- x}\} .$

Apply the concept of Wronskian,

$W [f_{1}, f_{2}, 鈥�, f_{n}] = |\begin{matrix} f_{1} (x) & f_{2} (x) & 鈥� & f_{n} (x) \\ f_{1}' (x) & f_{2}' (x) & 鈥� & f_{n}' (x) \\ 鈰� & 鈰� & 鈰� \\ {f_{1}}^{n - 1} (x) & {f_{2}}^{n - 1} (x) & 鈰� & {f_{n}}^{n - 1} (x) \end{matrix}|$

Therefore,

$W [e^{3 x}, e^{5 x}, e^{- x}] = |\begin{matrix} e^{3 x} & e^{5 x} & e^{- x} \\ 3 e^{3 x} & 5 e^{5 x} & - e^{- x} \\ 9 e^{3 x} & 25 e^{5 x} & e^{- x} \end{matrix}|$

Solve the above equation,

$W [e^{3 x}, e^{5 x}, e^{- x}] = e^{3 x} 脳 e^{5 x} 脳 e^{- x} |\begin{matrix} 1 & 1 & 1 \\ 3 & 5 & - 1 \\ 9 & 25 & 1 \end{matrix}| = e^{7 x} [1 (30) - 1 (12) + 1 (30)] = e^{7 x} [[48]] = 48 e^{7 x}$

02

Step 2:Check the linearly independent or dependent

Since the above result is $48 e^{7 x} 鈮� 0$ $(鈭赌 x)$ .

Therefore, $\{e^{3 x}, 鈥� e^{5 x}, 鈥� e^{- x}\}$ are linearly independent on $(- 鈭�, 鈥� 鈭�)$ .

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Most popular questions from this chapter

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

$y^{'''} - 2 y^{''} + y^{'} = x$

find a general solution to the given equation.

$y''' - 2 y'' - 5 y' + 6 y = e^{x} + x^{2}$

Find a general solution for the given linear system using the elimination method of Section 5.2.

$\begin{array}{l} \frac{d^{2} x}{d t^{2}} 鈭� x + 5 y = 0 \\ 2 x + \frac{d^{2} y}{d t^{2}} + 2 y = 0 \end{array}$

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.

$x \sqrt{x + 1} y''' - y' + xy = 0 y (\frac{1}{2}) = y' (\frac{1}{2}) = - 1, 鈥� 鈥� y'' (\frac{1}{2}) = 1$

find a differential operator that annihilates the given function.

$x^{2} e^{- x} s i n 2 x$

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