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ep 2 We have found that the roots of the auxiliary equation are m₁ = -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 yce+ce. Therefore, the complementary function is as follows. Yc-c₂e-x+c₂e-2x Let y₂e and y₂-e-2x be the two independent solutions which are terms of the complementary function. We will find functions u₂(x) and u₂(x) such that y-U₁₁+ ₂₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. w(x₂(x), ₂(x)) - w(ex, e-2) x₂(x) x₂(x) 6-2x

ep 2 We have found that the roots of the auxiliary equation are m₁ = -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 yce+ce. Therefore, the complementary function is as follows. Yc-c₂e-x+c₂e-2x Let y₂e and y₂-e-2x be the two independent solutions which are terms of the complementary function. We will find functions u₂(x) and u₂(x) such that y-U₁₁+ ₂₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. w(x₂(x), ₂(x)) - w(ex, e-2) x₂(x) x₂(x) 6-2x

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter1: Equations And Graphs
Section1.3: Lines
Problem 92E
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Step 2 We have found that the roots of the auxiliary equation are m₁ = -1 and m₂ = -2. In the case that the auxiliary equation for second-order linear equation has two distinct, real roots, we know the complementary function is of the form y = ₁₁x + ₂₂x. Therefore, the complementary function is as follows. Yc = C₁e-x + c₂e-2x Let y₁ = ex and y₂ = e-2x be the two independent solutions which are terms of the complementary function. We will find functions u₁(x) and u₂(x) such that y = U₁V₁ + U₂V₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), ₂(x)) = w(e-x, e-2x) |X₁ (X) X₂(X) | v₁'(x) x₂(x) A-X e-2x
Transcribed Image Text:Step 2 We have found that the roots of the auxiliary equation are m₁ = -1 and m₂ = -2. In the case that the auxiliary equation for second-order linear equation has two distinct, real roots, we know the complementary function is of the form y = ₁₁x + ₂₂x. Therefore, the complementary function is as follows. Yc = C₁e-x + c₂e-2x Let y₁ = ex and y₂ = e-2x be the two independent solutions which are terms of the complementary function. We will find functions u₁(x) and u₂(x) such that y = U₁V₁ + U₂V₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), ₂(x)) = w(e-x, e-2x) |X₁ (X) X₂(X) | v₁'(x) x₂(x) A-X e-2x
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