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Solve the differential equation by variation of parameters. y" + 3y + 2y We have found that the roots of the auxiliary equation are m₂ = -1 and m₂ = -2. In the case that the auxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementary function is of the form y=c₁₁x + ₂₂x. Therefore, the complementary function is as follows. Yc=c₂e-x + c²₂e-2x Let y₁=e* and Y₂ -2x be the two independent solutions which are terms of the complementary function. We will find functions u₂(x) and u₂(x) such that Vp = U₁V₁+U₂V₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), V₂(x)) = w(ex, e-2x) V1(x) V₂(x) |y₁'(x) v₂'(x)] =

Solve the differential equation by variation of parameters. y" + 3y + 2y We have found that the roots of the auxiliary equation are m₂ = -1 and m₂ = -2. In the case that the auxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementary function is of the form y=c₁₁x + ₂₂x. Therefore, the complementary function is as follows. Yc=c₂e-x + c²₂e-2x Let y₁=e* and Y₂ -2x be the two independent solutions which are terms of the complementary function. We will find functions u₂(x) and u₂(x) such that Vp = U₁V₁+U₂V₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), V₂(x)) = w(ex, e-2x) V1(x) V₂(x) |y₁'(x) v₂'(x)] =

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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Solve the differential equation by variation of parameters. y" + 3y + 2y We have found that the roots of the auxiliary equation are m₂ = -1 and m₂ = -2. In the case that the auxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementary function is of the form y=c₁₁x + ₂₂x. Therefore, the complementary function is as follows. Yc=c₂e-x + c²₂e-2x Let y₁=e* and Y₂ -2x be the two independent solutions which are terms of the complementary function. We will find functions u₂(x) and u₂(x) such that Vp = U₁V₁+U₂V₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), V₂(x)) = w(ex, e-2x) V1(x) V₂(x) |y₁'(x) v₂'(x)] =
Transcribed Image Text:Solve the differential equation by variation of parameters. y" + 3y + 2y We have found that the roots of the auxiliary equation are m₂ = -1 and m₂ = -2. In the case that the auxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementary function is of the form y=c₁₁x + ₂₂x. Therefore, the complementary function is as follows. Yc=c₂e-x + c²₂e-2x Let y₁=e* and Y₂ -2x be the two independent solutions which are terms of the complementary function. We will find functions u₂(x) and u₂(x) such that Vp = U₁V₁+U₂V₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), V₂(x)) = w(ex, e-2x) V1(x) V₂(x) |y₁'(x) v₂'(x)] =
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