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 y = ₁em ₁x + c₂em ₂x. 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 = U₁₁+U₂y₂ 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) Y₁(x) x₂(x) |y₁'(x) y₂'(x) e-x et e³r X e-2x -2e-2r
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 y = ₁em ₁x + c₂em ₂x. 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 = U₁₁+U₂y₂ 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) Y₁(x) x₂(x) |y₁'(x) y₂'(x) e-x et e³r X e-2x -2e-2r
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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Answer the wrong pls ty!
Transcribed Image Text: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 y = ₁₁x + ₂em 2x. Therefore, the complementary function is as follows.
с
e-x + c₂e-2x
Let y₁ = e-x 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 yµ = U₁Y₁ + ₂y 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)
Yc=c₁e
Ус
Y₁(x) Y₂(x)
|Y₁'(x) y/₂'(x)|
=
=
∙e
3.x
e
-X
e-x
X
e-2x
-2e-2r
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