Explanation Given: The initial value problem is y ″ − 2 x y ′ − 2 y = 0 ; y ( 0 ) = a 0 , y ′ ( 0 ) = a 1 , Here, a 0 and a 1 are constants. Approach: Rewrite the given equation a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 in the form y ″ + p ( x ) y ′ + q ( x ) y = 0 Here, p ( x ) = a 1 ( x ) a 2 ( x ) and q ( x ) = a 0 ( x ) a 2 ( x ) If x = 0 is an ordinary point for this equation then, the general solution of the given equation is y ( x ) = ∑ n = 0 ∞ a n x n . Calculation: The initial value problem is y ″ − 2 x y ′ − 2 y = 0 ... ( 1 ) Since p ( x ) = − 2 x and q ( x ) = − 2 are analytic at x = 0 therefore, write the general solution of the given problem as y ( x ) = ∑ n = 0 ∞ a n x n This gives y ′ ( x ) = ∑ n = 1 ∞ n a n x n − 1 y ″ ( x ) = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 Substitute these values in equation ( 1 ) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − 2 x ∑ n = 1 ∞ n a n x n − 1 − 2 ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − 2 ∑ n = 1 ∞ n a n x n − 2 ∑ n = 0 ∞ a n x n = 0 … ( 2 ) In order to make each term in the above equation like term, put k = n − 2 in the first summation, k = n in the second summation, k = n in the third summation Then, equation ( 2 ) becomes ∑ k = 0 ∞ ( k + 1 ) ( k + 2 ) a k + 2 x k − 2 ∑ k = 1 ∞ k a k x k − 2 ∑ k = 0 ∞ a k x k = 0 2 a 2 + ∑ k = 1 ∞ ( k + 1 ) ( k + 2 ) a k + 2 x k − 2 ∑ k = 1 ∞ k a k x k − 2 a 0 − 2 ∑ k = 1 ∞ a k x k = 0 2 ( a 2 − a 0 ) + ∑ k = 1 ∞ [ ( k + 1 ) ( k + 2 ) a k + 2 − 2 ( k + 1 ) a k ] x k = 0 This will make all the coefficients equal to zero To determine (b) To show: For positive values of a 0 and a 1 , the solution is increasing on ( 0 , ∞ ) . To determine (c) To find: For negative value of a 0 and positive value of a 1 , the solution is increasing on ( − ∞ , 0 ) . To determine (d) To find: The condition on a 0 and a 1 that guarantees that the solution is increasing on ( − ∞ , ∞ ) .

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 8.3, Problem 32E

To determine

(a)

To show:

For a0=0, the solution is an odd function and find the condition when a1=0.

To determine

(b)

To show:

For positive values of a0 and a1, the solution is increasing on (0,∞).

To determine

(c)

To find:

For negative value of a0 and positive value of a1, the solution is increasing on (−∞,0).

To determine

(d)

To find:

The condition on a0 and a1 that guarantees that the solution is increasing on (−∞,∞).

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.3, Problem 31E

Chapter 8.3, Problem 33E

Students have asked these similar questions

A cable television company estimates that with x thousand subscribers, its monthly revenue and cost (in thousands of dollars) are given by the following equations. R(x) = 45x - 0.24x2 C(x) = 257 + 13x

Compare the interest earned from #1 (where simple interest was used) to #5 (where compound interest was used). The principal, annual interest rate, and time were all the same; the only difference was that for #5, interest was compounded quarterly. Does the difference in interest earned make sense? Select one of the following statements. a. No, because more money should have been earned through simple interest than compound interest. b. Yes, because more money was earned through simple interest. For simple interest you earn interest on interest, not just on the amount of principal. c. No, because more money was earned through simple interest. For simple interest you earn interest on interest, not just on the amount of principal. d. Yes, because more money was earned when compounded quarterly. For compound interest you earn interest on interest, not just on the amount of principal.

Reduce the matrix to reduced row-echelon form. [3 2 -2-191 A = 3 -2 0 5 + 2 1 -2 -14 17 1 0 0 3 0 1 0 0 0 4

Chapter 8 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

(a) To show: For a 0 = 0 , the solution is an odd function and find the condition when a 1 = 0 .

Solution Summary: The author explains that the solution to a_0=0 is an odd function, and the required proof is given below.

Concept explainers

Videos

(a)

To show:

For a0=0, the solution is an odd function and find the condition when a1=0.

(b)

To show:

For positive values of a0 and a1, the solution is increasing on (0,∞).

(c)

To find:

For negative value of a0 and positive value of a1, the solution is increasing on (−∞,0).

(d)

To find:

The condition on a0 and a1 that guarantees that the solution is increasing on (−∞,∞).

Want to see the full answer?

Chapter 8 Solutions

(a) To show: For a 0 = 0 , the solution is an odd function and find the condition when a 1 = 0 .

Solution Summary: The author explains that the solution to a_0=0 is an odd function, and the required proof is given below.

Concept explainers

Videos

(a) To show: For a0=0, the solution is an odd function and find the condition when a1=0.

(b) To show: For positive values of a0 and a1, the solution is increasing on (0,∞).

(c) To find: For negative value of a0 and positive value of a1, the solution is increasing on (−∞,0).

(d) To find: The condition on a0 and a1 that guarantees that the solution is increasing on (−∞,∞).

Want to see the full answer?

Chapter 8 Solutions

(a)

To show:

For a0=0, the solution is an odd function and find the condition when a1=0.

(b)

To show:

For positive values of a0 and a1, the solution is increasing on (0,∞).

(c)

To find:

For negative value of a0 and positive value of a1, the solution is increasing on (−∞,0).

(d)

To find:

The condition on a0 and a1 that guarantees that the solution is increasing on (−∞,∞).