The General Solution The arbitrary linear combination y = C₁₁ + C₂y₂ of any two solutions with nonzero Wronskian is called the general solution. By the theorem we just proved, we can obtain any particular solution by adjusting C₁ and C₂. We now return to the solutions obtained earlier and prove the final result, for practical purposes the summarizing result. THEOREM 9.3.6 Given the equation y" + ay + by = 0, we form the characteristic equation r²+ ar+b= 0. I. If the characteristic equation has two distinct real roots r₁ and 12, then the general solution takes the form y= C₁e¹¹* + C₂ex II. If the characteristic equation has only one real root r = α, then the general solution takes the form y= C₁ex+C₂xe** = (C₁ + C₂x)e**. III. If the characteristic equation has two complex roots, r₁ = a + iẞ and then the general solution takes the form 12=α- iẞ, y= C₁ex cos ẞx + C₂ex sin ẞx = ex (C₁ cos ẞx+ C₂ sin ẞx). PROOF To prove this theorem, it is enough to show that the three solution pairs ex, ex ραν χρον ex cos ẞx, exsin ẞx all have nonzero Wronskians. The Wronskian of the first pair is the function d d W(x) = e^x(x) - (*) ex dx dx = ex ex-e^x ex = (12-11)(n+1)x W(x) is different from zero since, by assumption, 12 +11. We leave it to you to verify that the other pairs also have nonzero Wronskians. It is time to use what we have learned.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 31E
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Explain the key points of Theroem 9.3.6

The General Solution
The arbitrary linear combination y = C₁₁ + C₂y₂ of any two solutions with nonzero
Wronskian is called the general solution. By the theorem we just proved, we can obtain
any particular solution by adjusting C₁ and C₂.
We now return to the solutions obtained earlier and prove the final result, for
practical purposes the summarizing result.
THEOREM 9.3.6
Given the equation
y" + ay + by = 0,
we form the characteristic equation
r²+ ar+b= 0.
I. If the characteristic equation has two distinct real roots r₁ and 12, then the
general solution takes the form
y= C₁e¹¹* + C₂ex
II. If the characteristic equation has only one real root r = α, then the general
solution takes the form
y= C₁ex+C₂xe** = (C₁ + C₂x)e**.
III. If the characteristic equation has two complex roots,
r₁ = a + iẞ and
then the general solution takes the form
12=α- iẞ,
y= C₁ex cos ẞx + C₂ex sin ẞx = ex (C₁ cos ẞx+ C₂ sin ẞx).
Transcribed Image Text:The General Solution The arbitrary linear combination y = C₁₁ + C₂y₂ of any two solutions with nonzero Wronskian is called the general solution. By the theorem we just proved, we can obtain any particular solution by adjusting C₁ and C₂. We now return to the solutions obtained earlier and prove the final result, for practical purposes the summarizing result. THEOREM 9.3.6 Given the equation y" + ay + by = 0, we form the characteristic equation r²+ ar+b= 0. I. If the characteristic equation has two distinct real roots r₁ and 12, then the general solution takes the form y= C₁e¹¹* + C₂ex II. If the characteristic equation has only one real root r = α, then the general solution takes the form y= C₁ex+C₂xe** = (C₁ + C₂x)e**. III. If the characteristic equation has two complex roots, r₁ = a + iẞ and then the general solution takes the form 12=α- iẞ, y= C₁ex cos ẞx + C₂ex sin ẞx = ex (C₁ cos ẞx+ C₂ sin ẞx).
PROOF To prove this theorem, it is enough to show that the three solution pairs
ex, ex
ραν χρον
ex cos ẞx, exsin ẞx
all have nonzero Wronskians. The Wronskian of the first pair is the function
d
d
W(x) = e^x(x) - (*) ex
dx
dx
= ex ex-e^x ex
=
(12-11)(n+1)x
W(x) is different from zero since, by assumption, 12 +11.
We leave it to you to verify that the other pairs also have nonzero Wronskians.
It is time to use what we have learned.
Transcribed Image Text:PROOF To prove this theorem, it is enough to show that the three solution pairs ex, ex ραν χρον ex cos ẞx, exsin ẞx all have nonzero Wronskians. The Wronskian of the first pair is the function d d W(x) = e^x(x) - (*) ex dx dx = ex ex-e^x ex = (12-11)(n+1)x W(x) is different from zero since, by assumption, 12 +11. We leave it to you to verify that the other pairs also have nonzero Wronskians. It is time to use what we have learned.
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