How does Abel's Identity allow us to conclude that if the Wronskian is 0 at a point t, then the Wronskian must be identically 0
How does Abel's Identity allow us to conclude that if the Wronskian is 0 at a point t, then the Wronskian must be identically 0
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
How does Abel's Identity allow us to conclude that if the Wronskian is 0 at a point t, then the Wronskian must be identically 0
)'
W [y1, Y2](t)
(In |W[y1, 42](t)|)' = – P(t)
- P(t)
In |W[¥1 , 92](t)| = - | P(t)
W [y1, Y2](t) =Ce-S P(t)
Then we get what we want: W[y1, y2](t) = Ce-S P(t)
Note: To apply this to ay" + by' + cy = 0, divide by a to get y" + y + Ey = 0 and
let P = and Q = . Finally, you can solve for C by setting t = to in both sides of the
equation. That's how you get the more complicated formula:
W [y1, 42](t) = W[y1, Y2](to)e¯Sio P(»)ds](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F234a7be1-58c3-4913-b976-8fa32064a420%2F63395075-e98e-4a22-b11d-b0c37f972c03%2Fqzz702p_processed.png&w=3840&q=75)
Transcribed Image Text:PEYAM RYAN TABRIZIAN
(W[y1, y2](t))'
W [y1, Y2](t)
(In |W[y1, 42](t)|)' = – P(t)
- P(t)
In |W[¥1 , 92](t)| = - | P(t)
W [y1, Y2](t) =Ce-S P(t)
Then we get what we want: W[y1, y2](t) = Ce-S P(t)
Note: To apply this to ay" + by' + cy = 0, divide by a to get y" + y + Ey = 0 and
let P = and Q = . Finally, you can solve for C by setting t = to in both sides of the
equation. That's how you get the more complicated formula:
W [y1, 42](t) = W[y1, Y2](to)e¯Sio P(»)ds
 = Ce-S P()dt
Proof: This is actually MUCH easier than you think! First of all, by definition:
Y1 Y2
W[y1, Y2](t) =
Now differentiate:
(W [y1, y2](t))' = + 419% – y{y2 – = Y19% – y{y2
Now since y, and y2 both satisfy the differential equation above, we have:
yK + P(t)y% + Q(t)yı = 0
y = -P(t)y – Q(t)y1
Y% + P(t)y2 + Q(t)y2 = 0
Y% = -P(t)y½ – Q(t)y2
Hence:
Y14% – yiy2 =y1 (-P(t)y½ – Q(t)y2) – (-P(t)y½ – Q(t)yı) Y2
=- P(t)y12 - Q9TT2 + P(t)¾y2 + Q(H0n2
:- P(t) (y12 – i42)
But this is just equal to – P(t)W[y1, Y2](t), hence:
(W [y1, y2](t))' = –P(t)W[y1,Y2](t)
All we need to do is to solve this differential equation, but this is very similar to Math
1B:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F234a7be1-58c3-4913-b976-8fa32064a420%2F63395075-e98e-4a22-b11d-b0c37f972c03%2Fd7x9ami_processed.png&w=3840&q=75)
Transcribed Image Text:ABEL'S FORMULA
PEYAM RYAN TABRIZIAN
Abel's formula: Suppose y" + P(t)y' + Q(t)y = 0. Then:
W [y1, y2](t) = Ce-S P()dt
Proof: This is actually MUCH easier than you think! First of all, by definition:
Y1 Y2
W[y1, Y2](t) =
Now differentiate:
(W [y1, y2](t))' = + 419% – y{y2 – = Y19% – y{y2
Now since y, and y2 both satisfy the differential equation above, we have:
yK + P(t)y% + Q(t)yı = 0
y = -P(t)y – Q(t)y1
Y% + P(t)y2 + Q(t)y2 = 0
Y% = -P(t)y½ – Q(t)y2
Hence:
Y14% – yiy2 =y1 (-P(t)y½ – Q(t)y2) – (-P(t)y½ – Q(t)yı) Y2
=- P(t)y12 - Q9TT2 + P(t)¾y2 + Q(H0n2
:- P(t) (y12 – i42)
But this is just equal to – P(t)W[y1, Y2](t), hence:
(W [y1, y2](t))' = –P(t)W[y1,Y2](t)
All we need to do is to solve this differential equation, but this is very similar to Math
1B:
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By Abel's identity conclude that if Wronskian is 0 at a point , then the Wronskian is identically equal to 0.
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