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Find a fundamental set of solutions for y" - 4y' + 4y = 0.

Find a fundamental set of solutions for y" - 4y' + 4y = 0.

Algebra for College Students
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter10: Exponential And Logarithmic Functions
Section10.CR: Review Problem Set
Problem 60CR
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Find a fundamental set of solutions for y" - 4y' + 4y = 0.

 

=e²t, The characteristic equation, r² −4r +4=0, factors as (r−2)² = 0, so its two roots are r₁=₂=2. One solution is y₁ = e but it is impossible to also have y₂ = e²t because then W[₁,₂]=0. (Verify this: W[e²,e²¹]=[]; explain why it is a 2t 2t problem: ) By the Principle of Superposition, we know cy₁ (t) is a solution for any value of the constant c. Are there any more solutions? Maybe in the form y₂ =v(t)y₁ (t)=v(t)e² for some function v(t)? To check, use the Product Rule to calculate y₂ Y₂"-4y₂' +4y₂ So, all of this simplifies to v" = V= - and Y₂": = -4( . One integration gives v' = for any constants c₂ and " for any C₂. There are lots of solutions: y=v(t)e²t 2t C₂. One of these two terms is just y₁=e²t, but the other term is the second solution: y₂ = constants c₁ and check that {₁,₂} is a fundamental set of solutions for y" −4y'+4y=0, compute their Wronskian: W[₁,₂]= . Then ● ])+4(ve²) = (v")e²t = 0.. and a second integration provides =C₁ +C₂ . To
Transcribed Image Text:=e²t, The characteristic equation, r² −4r +4=0, factors as (r−2)² = 0, so its two roots are r₁=₂=2. One solution is y₁ = e but it is impossible to also have y₂ = e²t because then W[₁,₂]=0. (Verify this: W[e²,e²¹]=[]; explain why it is a 2t 2t problem: ) By the Principle of Superposition, we know cy₁ (t) is a solution for any value of the constant c. Are there any more solutions? Maybe in the form y₂ =v(t)y₁ (t)=v(t)e² for some function v(t)? To check, use the Product Rule to calculate y₂ Y₂"-4y₂' +4y₂ So, all of this simplifies to v" = V= - and Y₂": = -4( . One integration gives v' = for any constants c₂ and " for any C₂. There are lots of solutions: y=v(t)e²t 2t C₂. One of these two terms is just y₁=e²t, but the other term is the second solution: y₂ = constants c₁ and check that {₁,₂} is a fundamental set of solutions for y" −4y'+4y=0, compute their Wronskian: W[₁,₂]= . Then ● ])+4(ve²) = (v")e²t = 0.. and a second integration provides =C₁ +C₂ . To
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