3. y4) + 4y" = 0; y(0) = 0, y'(0) = -1, y₁ (t) = 1, y₂ (t)=t, y3(t) = cos 2t, y'(0)=-4, y" (0) = 8 y4 (t) = sin 2t
3. y4) + 4y" = 0; y(0) = 0, y'(0) = -1, y₁ (t) = 1, y₂ (t)=t, y3(t) = cos 2t, y'(0)=-4, y" (0) = 8 y4 (t) = sin 2t
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
Please show all work and do all parts. Only do question 3.
Transcribed Image Text:aten
Second and Higher Order Linear Differential Equations
2. y"" - y' = 0; y(0) = 4, y'(0) = 1,
y₁ (t) = 1, y₂ (t)=e¹, y(t) = et
13. y(4) + 4y" = 0; y(0) = 0, y'(0) = -1,
y₁ (t) = 1, y₂ (t)=t, y3 (t) = cos 2t,
4. y" + 2y" = 0; y(0) = 0, y'(0) = 3,
y₁ (t) = 1, y₂(t)=t, y3(t)=e-²t
5. ty" + 3y"=0, t>0;
y₂(t) =t,
y₁ (t) = 1,
6. ty" + ty" - y = 0,
Exercises 7-10:
y"(0) = 3
y₁ (t) = 1, y₂ (t) = ln(-t),
y(2) = 1, y' (2)=-,
y3(t) = t¹3.00
WOH
y'(0) = -4, y" (0) = 8
y4 (t) = sin 2t
03 ol
y" (0) = -8
y" (2) =
sorod sepgil ba
t<0; y(-1) = 1, y'(-1) = -1, y'(-1) =
y3 (t) = t²
Consider the given differential equation on the interval -∞<t<∞..
members of a solution set satisfy the initial conditions. Do the solution
mental set?
7. y" + 2ty +1²y = 0, y₁ (1)=2, ₁(1) = -1, 3₂(1) = -4, y/2(1)
(0)
CO
![S
only if the determinant of A is nonzero.
• PROOF: We prove Theorem 3.9 for the case n = 2. Since (y₁, y2) is a
damental set, we can express y₁ and 2 as linear combinations of y, and y2
y₁ = a11y1 + a21Y2
2 = a121 + a2232
or [1,2] [Y₁, Y₂]
Since the equation [₁,2] = [y₁, y2lA holds for all t in (a, b), we can different
it and obtain the matrix equation
V₁ V2
[₁₂]
1. y"" = 0;
=
y₁ (t) = 2,
31 32
[V₁ Y2]
[a₁1 912)
a21 a22
Using the fact that the determinant of the product of two matrices is the pro
of their determinants, we have
= [y₁, y2]A.
A.
W(t) = W(t) det(A),
1:
where W (t) and W(t) denote the Wronskians of {₁, 2} and {y₁, y2), respecti
Since W (t) #0 on (a, b), it follows that W(t) # 0 if and only if det(A) # 0.
Exercises 1-6:
In each exercise,
(a) Verify that the given functions form a fundamental set of solutions.
(b) Solve the initial value problem.
y(1) = 4, y'(1) = 2, y" (1) = 0
y₂(t) = t-1, y3(t) = t² - 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71b114d5-6d72-4063-8215-3d54ecc2c41a%2Fe63a2692-7ab9-442e-a933-109942b05531%2F6zu9efe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:S
only if the determinant of A is nonzero.
• PROOF: We prove Theorem 3.9 for the case n = 2. Since (y₁, y2) is a
damental set, we can express y₁ and 2 as linear combinations of y, and y2
y₁ = a11y1 + a21Y2
2 = a121 + a2232
or [1,2] [Y₁, Y₂]
Since the equation [₁,2] = [y₁, y2lA holds for all t in (a, b), we can different
it and obtain the matrix equation
V₁ V2
[₁₂]
1. y"" = 0;
=
y₁ (t) = 2,
31 32
[V₁ Y2]
[a₁1 912)
a21 a22
Using the fact that the determinant of the product of two matrices is the pro
of their determinants, we have
= [y₁, y2]A.
A.
W(t) = W(t) det(A),
1:
where W (t) and W(t) denote the Wronskians of {₁, 2} and {y₁, y2), respecti
Since W (t) #0 on (a, b), it follows that W(t) # 0 if and only if det(A) # 0.
Exercises 1-6:
In each exercise,
(a) Verify that the given functions form a fundamental set of solutions.
(b) Solve the initial value problem.
y(1) = 4, y'(1) = 2, y" (1) = 0
y₂(t) = t-1, y3(t) = t² - 1
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