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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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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