Find a fundamental set of solutions for y" - 4y' + 4y = 0.
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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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe00cebfb-aec4-474d-8979-79a2d105b819%2Fcaa9c93d-c7f6-47fc-a755-fd95bccd1953%2F4jwc85y_processed.jpeg&w=3840&q=75)
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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