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dyex Example// Solve = ex-Y dx Solution/ dy dy + Y = ex >> dx + p(x)y = Q(x) dx p(x) = 1 & Q(x) = ex p = e√ p(x)dx = e√ 1dx = ex exdx y = e−√(x)dx + ( p + Q(x) + c)=e^x + (ex + e* dx + c) = * C * 1 * e¯x, (√ e²x dx + c) = e²¯x ( ²²x + c) = ½³+exc *

dyex Example// Solve = ex-Y dx Solution/ dy dy + Y = ex >> dx + p(x)y = Q(x) dx p(x) = 1 & Q(x) = ex p = e√ p(x)dx = e√ 1dx = ex exdx y = e−√(x)dx + ( p + Q(x) + c)=e^x + (ex + e* dx + c) = * C * 1 * e¯x, (√ e²x dx + c) = e²¯x ( ²²x + c) = ½³+exc *

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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dyex Example// Solve = ex-Y dx Solution/ dy dy + Y = ex >> dx + p(x)y = Q(x) dx p(x) = 1 & Q(x) = ex p = e√ p(x)dx = e√ 1dx = ex exdx y = e−√(x)dx + ( p + Q(x) + c)=e^x + (ex + e* dx + c) = * C * 1 * e¯x, (√ e²x dx + c) = e²¯x ( ²²x + c) = ½³+exc *
Transcribed Image Text:dyex Example// Solve = ex-Y dx Solution/ dy dy + Y = ex >> dx + p(x)y = Q(x) dx p(x) = 1 & Q(x) = ex p = e√ p(x)dx = e√ 1dx = ex exdx y = e−√(x)dx + ( p + Q(x) + c)=e^x + (ex + e* dx + c) = * C * 1 * e¯x, (√ e²x dx + c) = e²¯x ( ²²x + c) = ½³+exc *
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