Divya sree reddy Asked a Question
July 12, 2020 3:16 pmpts 30 pts
Let y: R > R satisfy the initial value problem y(t) = 1 y(t) where t e R and y(0) = 0. Then which of the following is correct ? (A) y(t,) = 1 for somet, e R. y(t)> - 1 for all t e R. (B) (C)yt) is strictly decreasing inR. (D) y is increasing in (0, 1) & decreasing in (1, o).
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  • Ujjawal vishal thankyou
    option b see attached
    • cropped4531110829225566486.jpg
    • cropped3761199695973025816.jpg
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    Divya sree reddy
    here I didn't get the fourth step, I think it should be , log((1+y)(1-y)). Can you please explain?
  • Divya sree reddy
    Once please check the last step in the image.
    • cropped954010482.jpg
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  • Divya sree reddy
    can we solve in this way by using partial fractions?
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