of the form y₁ = (1+₁2+ a₂x² + 3x³ + ...) 32 = x2(1+b₁x + b₂x² + b₂x³ + ...) where T₁ > T2- Enter Find two linearly independent solutions of 2x²y" - xy + (2x+1)y=0, z>0 T1 1 a₁ = 9₂= 03 = T2 1/2 b₁ = b₂ == b3 =
of the form y₁ = (1+₁2+ a₂x² + 3x³ + ...) 32 = x2(1+b₁x + b₂x² + b₂x³ + ...) where T₁ > T2- Enter Find two linearly independent solutions of 2x²y" - xy + (2x+1)y=0, z>0 T1 1 a₁ = 9₂= 03 = T2 1/2 b₁ = b₂ == b3 =
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 26E
Related questions
Question
![of the form
y₁ =
(1+₁+ a₂²+az³ + ...)
3₂ = x2(1+b₁x + b₂x² + b₂x³ + ...)
where T₁ > T2-
Enter
Find two linearly independent solutions of 2x²y" - xy + (2x+1)y=0, z>0
T1 1
01 =
a₂=
03 =
T2= 1/2
b₁ =
b₂ =
b3 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7257e379-412a-45af-8489-b06fafcd19b9%2F3b26a114-a9f5-48aa-9a2d-645e448742ed%2Fvyf4u4c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:of the form
y₁ =
(1+₁+ a₂²+az³ + ...)
3₂ = x2(1+b₁x + b₂x² + b₂x³ + ...)
where T₁ > T2-
Enter
Find two linearly independent solutions of 2x²y" - xy + (2x+1)y=0, z>0
T1 1
01 =
a₂=
03 =
T2= 1/2
b₁ =
b₂ =
b3 =
Expert Solution
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Step 1: We define a regular singular point of a second order differential equation.
VIEWStep 2: We write the solution in series form using Frobenius method.
VIEWStep 3: Calculating y(x)
VIEWStep 4: Finding indicial equation.
VIEWStep 5: Finding the coefficients.
VIEWStep 6: Write y(x) in k and x
VIEWStep 7: Finding two linearly independent solutions.
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