Verify that both_y₁(t) = 1-t_and_y₂(t) = -t²/4 are solutions of the initial value problem = -t+ 1²+ 4y 2 C₁ d. e- y(2)= Where are these solutions valid? qy₁(t) is a solution for b. y₁(t) is a solution for = y₁(t) is a solution for y₁(t) is a solution for y₁(t) is a solution for - 1 t≥ 2, t> -2 y₂(t) is a solution of all t y₂(t) is a solution for t≥ 0 t > -2 y₂(t) is a solution of all t t≥2, y₂(t) is a solution for t≥0 t≤2, y₂(t) is a solution of all t "

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Verify that both y₁(t) = 1−t and y₂(t) = −t²/4 are solutions of the initial value problem
y' = = ¹+ √²
2
Where are these solutions valid?
-t+ t² +4y
y(2) =
=
-1
qy₁(t) is a solution for t > 2
2
b. y₁(t) is a solution for
Cy₁(t)
d
e-
y₂(t) is a solution of all t
Y₂(t) is a solution for t≥0
is a solution for
Y₂(t) is a solution of all t
y₁(t) is a solution for
t≥2,
y₂(t) is a solution for t > 0
y₁(t) is a solution for t≤2, y₂(t) is a solution of all t
t> -2
t> -2
2
>
Transcribed Image Text:Verify that both y₁(t) = 1−t and y₂(t) = −t²/4 are solutions of the initial value problem y' = = ¹+ √² 2 Where are these solutions valid? -t+ t² +4y y(2) = = -1 qy₁(t) is a solution for t > 2 2 b. y₁(t) is a solution for Cy₁(t) d e- y₂(t) is a solution of all t Y₂(t) is a solution for t≥0 is a solution for Y₂(t) is a solution of all t y₁(t) is a solution for t≥2, y₂(t) is a solution for t > 0 y₁(t) is a solution for t≤2, y₂(t) is a solution of all t t> -2 t> -2 2 >
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