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Step 1 We are given a nonhomogeneous second-order differential equation. Similar to the method of solving by undetermined coefficients, we first find the complementary function y for the associated homogeneous equation. This time, the particular solution is based on Wronskian determinants and the general solution is y = y + yp First, we must find the roots of the auxiliary equation for y" + 3y + 2y = 0. m² +3m +2=0 Solving for m, the roots of the auxiliary equation are as follows. smaller value larger value my m₂ |y₁(x) x₂(x) X₁'(X) X₂'(x) | = 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 a second-order linear equation has two distinct, real roots, we know the complementary function is of the form Yc=c₁e₁x + ₂₂x. Therefore, the complementary function is as follows. C. -2x Yc=c₁ex + c₂e-2 -2 Let y 1 = e 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 = U₁₁+U₂₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. 2 W(y₁(x), ₂(x)) = W(e-x, e-2x) = ex e-2x

Step 1 We are given a nonhomogeneous second-order differential equation. Similar to the method of solving by undetermined coefficients, we first find the complementary function y for the associated homogeneous equation. This time, the particular solution is based on Wronskian determinants and the general solution is y = y + yp First, we must find the roots of the auxiliary equation for y" + 3y + 2y = 0. m² +3m +2=0 Solving for m, the roots of the auxiliary equation are as follows. smaller value larger value my m₂ |y₁(x) x₂(x) X₁'(X) X₂'(x) | = 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 a second-order linear equation has two distinct, real roots, we know the complementary function is of the form Yc=c₁e₁x + ₂₂x. Therefore, the complementary function is as follows. C. -2x Yc=c₁ex + c₂e-2 -2 Let y 1 = e 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 = U₁₁+U₂₂ is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. 2 W(y₁(x), ₂(x)) = W(e-x, e-2x) = ex e-2x

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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Step 1 We are given a nonhomogeneous second-order differential equation. Similar to the method of solving by undetermined coefficients, we first find the complementary function y for the associated homogeneous equation. This time, the particular solution y is based on Wronskian determinants and the general solution is y = y + Yp² First, we must find the roots of the auxiliary equation for y" + 3y² + 2y = 0. m² +3m + 2 = 0 Solving for m, the roots of the auxiliary equation are as follows. smaller value larger value m₁ X₁(x) x₂(x) V₁'(x) V₂'(x) m₂ = = = 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 a second-order linear equation has two distinct, real roots, we know the complementary function is of the form Yc=c₁₁x + c₂em 2x. Therefore, the complementary function is as follows. YcC₁ex + c₂e-2x -2x Let y₁ = ex and y₂ = e be the two independent solutions which are terms of the complementary function. We will find functions u₁(x) and u₂(x) such that yp = U₁Y₁+U₂2 is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), y₂(x)) = w(e-x, e-2x) -2 e-x e-2x
Transcribed Image Text:Step 1 We are given a nonhomogeneous second-order differential equation. Similar to the method of solving by undetermined coefficients, we first find the complementary function y for the associated homogeneous equation. This time, the particular solution y is based on Wronskian determinants and the general solution is y = y + Yp² First, we must find the roots of the auxiliary equation for y" + 3y² + 2y = 0. m² +3m + 2 = 0 Solving for m, the roots of the auxiliary equation are as follows. smaller value larger value m₁ X₁(x) x₂(x) V₁'(x) V₂'(x) m₂ = = = 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 a second-order linear equation has two distinct, real roots, we know the complementary function is of the form Yc=c₁₁x + c₂em 2x. Therefore, the complementary function is as follows. YcC₁ex + c₂e-2x -2x Let y₁ = ex and y₂ = e be the two independent solutions which are terms of the complementary function. We will find functions u₁(x) and u₂(x) such that yp = U₁Y₁+U₂2 is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y₁(x), y₂(x)) = w(e-x, e-2x) -2 e-x e-2x
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