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It is known that the two basic solutions of y" +b (x equation y" + b (x) y' +c(x) y = f (x) we use w arctan(x). So this is the COMPLETE solu %3D and u2 = O y = C1e + C2xe² – In(1 + ²) + arctan(x) In (1 + x?) + arctan(x) | O y = Cie + C2xe² - e In(1 + x²) + xe² arctan O y = -e In(1+z²) + xe² arctan(x) %3D O y = Cje? + C2xe² + e²/1 – x² + xe² arcsin(x) %3D

It is known that the two basic solutions of y" +b (x equation y" + b (x) y' +c(x) y = f (x) we use w arctan(x). So this is the COMPLETE solu %3D and u2 = O y = C1e + C2xe² – In(1 + ²) + arctan(x) In (1 + x?) + arctan(x) | O y = Cie + C2xe² - e In(1 + x²) + xe² arctan O y = -e In(1+z²) + xe² arctan(x) %3D O y = Cje? + C2xe² + e²/1 – x² + xe² arcsin(x) %3D

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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It is known that the two basic solutions of y" +b (x) y' + c (x) y = 0 are y1 = equation y" +b(x) y' + c (x) y= f (x) we use Variation of Parameters to find that u1 = arctan(x). So this is the COMPLETE solution of y" + b (x) y' + c(x)y= f (x). et and y2 = xe. For the %3D - -1n(1 + 교2) and u2 = O y = Cie + C2xe² – In(1 + x²) + arctan(r) O y = Cie + C2xe² - e In(1+ x?) + xe² arctan(r) O y = -e In(1+z²) + xe² arctan(x) O y = C1e? + C2xe² + e²/1 – a² + xe² arcsin(x)
Transcribed Image Text:It is known that the two basic solutions of y" +b (x) y' + c (x) y = 0 are y1 = equation y" +b(x) y' + c (x) y= f (x) we use Variation of Parameters to find that u1 = arctan(x). So this is the COMPLETE solution of y" + b (x) y' + c(x)y= f (x). et and y2 = xe. For the %3D - -1n(1 + 교2) and u2 = O y = Cie + C2xe² – In(1 + x²) + arctan(r) O y = Cie + C2xe² - e In(1+ x?) + xe² arctan(r) O y = -e In(1+z²) + xe² arctan(x) O y = C1e? + C2xe² + e²/1 – a² + xe² arcsin(x)
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