Question

a. Verify that both y1(t) = 1 − t and y2(t) = −(t^2)/4 are solutions of the initial value problem y' = ((−t )+ ((t^2) +4y)^(1/2))/2 , y(2) = − 1 Where are these solutions valid? b. Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of the existence and uniqueness for non-linear equations theorem.

Name: a verify that both y1t 1 t and y2t t24 are solutions of the initial value problem y t t2 4y122 y2 1 where are these solutions valid b explain why the existence of two solutions of the given 14601
Uploaded: 2022-10-10T12:03:15-08:00
Duration: 4 min 7 s
Channel: Sri K
Description: a verify that both y1t 1 t and y2t t24 are solutions of the initial value problem y t t2 4y122 y2 1 where are these solutions valid b explain why the existence of two solutions of the given 14601

          a. Verify that both y1(t) = 1 − t and y2(t) = −(t^2)/4 are solutions of the initial value problem y' = ((−t )+ ((t^2) +4y)^(1/2))/2 , y(2) = − 1
    Where are these solutions valid?
    b. Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of the existence and uniqueness for non-linear equations theorem.

Added by Mark W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

10/10/2022

Step 1

Substitute y1(t) = 1 - t into the differential equation: y' = ((-t) + (t^2 + 4(1 - t))^(1/2))/2 y' = ((-t) + (t^2 + 4 - 4t)^(1/2))/2 y' = ((-t) + (t^2 - 4t + 4)^(1/2))/2 y' = ((-t) + (t - 2)^2)^(1/2))/2 y' = ((-t + t - 2))/2 y' = -1 Now, substitute y(2) = -1 Show more…

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a. Verify that both y1(t) = 1 − t and y2(t) = −(t^2)/4 are solutions of the initial value problem y' = ((−t )+ ((t^2) +4y)^(1/2))/2 , y(2) = − 1 Where are these solutions valid? b. Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of the existence and uniqueness for non-linear equations theorem.