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3. Verify that the following functions are solution of one of the equations in part 1 and match the solution to its equation. а. 2е3x b. e* + x + 2 c. e3x + x + 2

3. Verify that the following functions are solution of one of the equations in part 1 and match the solution to its equation. а. 2е3x b. e* + x + 2 c. e3x + x + 2

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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3. Verify that the following functions are solution of one of the equations in part 1 and match the solution to its equation. a. 2e3x b. e* + x + 2 с. езх + х + 2
Transcribed Image Text:3. Verify that the following functions are solution of one of the equations in part 1 and match the solution to its equation. a. 2e3x b. e* + x + 2 с. езх + х + 2
Example 1. Show that y(x) = x 3/2 is a solution to 4x?y" + 12xy' + 3y = 0 for x > 0 Solution: Since the highest order given D.E. is two, we'll need to do is to get the first and second derivative respectively of y(x) = x 3/2 3 15 y (2) = - y" (æ) = 2 Substituting these derivatives into the differential equation, we have: 3 4x2 ?x 7/2) + 12x (–x 5/2) + 3 (x -32) = 0 2 -3/2 15x 2 – 18x -3/2 + 3x 3/2 = 0 0 = 0; Ok! Example 2. Verify that y = e¯2x is a solution to the equation y" – 3y' + 2y = 0 Solution: Since the highest order given D.E. is three, we'll need to do is to get the first, second and third derivative respectively of y = e -2x y = - 2e-2x y" = 4e 2x = - 8e 2x Substituting these derivatives into the differential equation, we have: - 8e-2x - 3 (- 2e2×) + 2(e2×) = 0 - 8e 2x + 6e2x + 2e 2 = 0 8e 2x + 8e 2x = 0 0 = 0, Ok!
Transcribed Image Text:Example 1. Show that y(x) = x 3/2 is a solution to 4x?y" + 12xy' + 3y = 0 for x > 0 Solution: Since the highest order given D.E. is two, we'll need to do is to get the first and second derivative respectively of y(x) = x 3/2 3 15 y (2) = - y" (æ) = 2 Substituting these derivatives into the differential equation, we have: 3 4x2 ?x 7/2) + 12x (–x 5/2) + 3 (x -32) = 0 2 -3/2 15x 2 – 18x -3/2 + 3x 3/2 = 0 0 = 0; Ok! Example 2. Verify that y = e¯2x is a solution to the equation y" – 3y' + 2y = 0 Solution: Since the highest order given D.E. is three, we'll need to do is to get the first, second and third derivative respectively of y = e -2x y = - 2e-2x y" = 4e 2x = - 8e 2x Substituting these derivatives into the differential equation, we have: - 8e-2x - 3 (- 2e2×) + 2(e2×) = 0 - 8e 2x + 6e2x + 2e 2 = 0 8e 2x + 8e 2x = 0 0 = 0, Ok!
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