x³ y" + xy' - y = 0 [ Hint: x = e¹, ln x = t; x(dy/dx) = dy/dt; x²(d²y/dx²) x³(d³y/dx³) = (d²y/dt2) - dy/dt; = (d³y/dt³) - 3(d²y/dt²) + 2(dy/dt)]
x³ y" + xy' - y = 0 [ Hint: x = e¹, ln x = t; x(dy/dx) = dy/dt; x²(d²y/dx²) x³(d³y/dx³) = (d²y/dt2) - dy/dt; = (d³y/dt³) - 3(d²y/dt²) + 2(dy/dt)]
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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