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y" (x) = 2y(x) + 2[y(x)]° with y(0) = 0 and y'(0) = 1.

y" (x) = 2y(x) + 2[y(x)]° with y(0) = 0 and y'(0) = 1.

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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Consider y" = f(y) ; by multiplying by y' (x) both the left and the right hand sides can be swiftly integrated as 1 (y')² = F(y) + C where dFldy = f(y). y" (x) = [y(x)]² can be rewritten as: So, for example, 1 1 Ly(x)]° + C. We will use this trick on another simple case with an exact integral. 6. Use the technique above to find y(x) such that y"(x) = 2y(x) + 2[y(x)]³ with y(0) = 0 and y' (0) = 1. Hint: Once you use the above to simplify the expression to the form y' = g(y), you can solve it by moving g(y) into the denominator: ly/g(y) = dx
Transcribed Image Text:Consider y" = f(y) ; by multiplying by y' (x) both the left and the right hand sides can be swiftly integrated as 1 (y')² = F(y) + C where dFldy = f(y). y" (x) = [y(x)]² can be rewritten as: So, for example, 1 1 Ly(x)]° + C. We will use this trick on another simple case with an exact integral. 6. Use the technique above to find y(x) such that y"(x) = 2y(x) + 2[y(x)]³ with y(0) = 0 and y' (0) = 1. Hint: Once you use the above to simplify the expression to the form y' = g(y), you can solve it by moving g(y) into the denominator: ly/g(y) = dx
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