Which of the following differential equations with initial condition yields a solution curve that exhibits modified exponential growth such that y goes to negative infinity as t goes to positive infinity? (Hint: Check each one by solving the differential equation quickly. Remember, to solve the differential equation the quick and easy way, recall the fact that the solution to the differential equation=k(-a) is y = Aekt + a. So you just need to write the differential equation in the proper form first. Then, of course, you must solve for the constant A using the initial condition.) OA.=2(+200) with initial condition y(0) = 50 B. = 2(y-200) with initial condition y(0) = 50 dt O C. = 3.5(+123) with initial condition (0) = 50 O D. = -20-200) with initial condition y(0) = 100 O E. E. dy = -3.5(+123) with initial condition (0) - 100

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Question
Which of the following differential equations with initial condition yields a
solution curve that exhibits modified exponential growth such that y goes to
negative infinity as tgoes to positive infinity?
(Hint: Check each one by solving the differential equation quickly. Remember,
to solve the differential equation the quick and easy way, recall the fact that
the solution to the differential equation - ky -a) is y = Aekt +a.
So you just need to write the differential equation in the proper form first.
Then, of course, you must solve for the constant A using the initial condition.)
А.
2(7 + 200) with initial condition p(0) = 50
B. - 2(y - 200) with initial condition y(0) = 50
O c. = 3.5(y + 123) with initial condition y(0) = 50
dt
D.
dy
--2 - 200) with initial condition y(0) = 100
dy
OE.
=-3.5(y + 123) with initial condition y(0) = 100
Transcribed Image Text:Which of the following differential equations with initial condition yields a solution curve that exhibits modified exponential growth such that y goes to negative infinity as tgoes to positive infinity? (Hint: Check each one by solving the differential equation quickly. Remember, to solve the differential equation the quick and easy way, recall the fact that the solution to the differential equation - ky -a) is y = Aekt +a. So you just need to write the differential equation in the proper form first. Then, of course, you must solve for the constant A using the initial condition.) А. 2(7 + 200) with initial condition p(0) = 50 B. - 2(y - 200) with initial condition y(0) = 50 O c. = 3.5(y + 123) with initial condition y(0) = 50 dt D. dy --2 - 200) with initial condition y(0) = 100 dy OE. =-3.5(y + 123) with initial condition y(0) = 100
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