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Problem 3: In the following problems, there is an argument done incorrectly. For each part, state the location of the error, and provide a one sentence description of the mistake. Ce™ where C > 0. (Note this claim is false, since -e" (b) This is a proof that if y' = y, then y(x) satisfies, so there must be a mistake.) (i) The equation y' = y is separable, so we divide y' = y on both sides by y to get: y'(x) y(x) ' (ii) Note that: y' (x) y(x) In(y(x)) + 1= da In(y(x)). dx (iii) Integrate both sides, to find: x +c = In(y(x)). (iv) Take exp of both sides: er+c = y(x). (v) Note exp(x + c) = exp(x) exp(c). Let C = exp(C), and we get: C'e = y(x). (vi) Since exp(c) > 0 for all c e R, we conclude C > 0.

Problem 3: In the following problems, there is an argument done incorrectly. For each part, state the location of the error, and provide a one sentence description of the mistake. Ce™ where C > 0. (Note this claim is false, since -e" (b) This is a proof that if y' = y, then y(x) satisfies, so there must be a mistake.) (i) The equation y' = y is separable, so we divide y' = y on both sides by y to get: y'(x) y(x) ' (ii) Note that: y' (x) y(x) In(y(x)) + 1= da In(y(x)). dx (iii) Integrate both sides, to find: x +c = In(y(x)). (iv) Take exp of both sides: er+c = y(x). (v) Note exp(x + c) = exp(x) exp(c). Let C = exp(C), and we get: C'e = y(x). (vi) Since exp(c) > 0 for all c e R, we conclude C > 0.

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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Problem 3: In the following problems, there is an argument done incorrectly. For each part, state the location of the error, and provide a one sentence description of the mistake. Ce™ where C > 0. (Note this claim is false, since -e" (b) This is a proof that if y' = y, then y(x) satisfies, so there must be a mistake.) (i) The equation y' = y is separable, so we divide y' = y on both sides by y to get: y'(x) y(x) ' (ii) Note that: y' (x) y(x) In(y(x)) + 1= da In(y(x)). dx (iii) Integrate both sides, to find: x +c = In(y(x)). (iv) Take exp of both sides: er+c = y(x). (v) Note exp(x + c) = exp(x) exp(c). Let C = exp(C), and we get: C'e = y(x). (vi) Since exp(c) > 0 for all c e R, we conclude C > 0.
Transcribed Image Text:Problem 3: In the following problems, there is an argument done incorrectly. For each part, state the location of the error, and provide a one sentence description of the mistake. Ce™ where C > 0. (Note this claim is false, since -e" (b) This is a proof that if y' = y, then y(x) satisfies, so there must be a mistake.) (i) The equation y' = y is separable, so we divide y' = y on both sides by y to get: y'(x) y(x) ' (ii) Note that: y' (x) y(x) In(y(x)) + 1= da In(y(x)). dx (iii) Integrate both sides, to find: x +c = In(y(x)). (iv) Take exp of both sides: er+c = y(x). (v) Note exp(x + c) = exp(x) exp(c). Let C = exp(C), and we get: C'e = y(x). (vi) Since exp(c) > 0 for all c e R, we conclude C > 0.
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