solve y’+(sinx)y=0 with intial value y(π/2)=1​

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solve y’+(sinx)y=0 with intial value y(π/2)=1​

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Eliza 1 month 2021-08-12T09:10:25+00:00 1 Answer 0 views 0

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    0
    2021-08-12T09:11:39+00:00

    Step-by-step explanation:

    So long as y is not 25, we can rewrite the differential equation as

    dy

    dt

    1

    25 − y

    = 2

    1

    25 − y

    dy = 2 dt,

    so

    Z

    1

    25 − y

    dy =

    Z

    2 dt,

    that is, the two anti-derivatives must be the same except for a constant difference. We can

    calculate these anti-derivatives and rearrange the results:

    Z

    1

    25 − y

    dy =

    Z

    2 dt

    (−1) ln |25 − y| = 2t + C0

    ln |25 − y| = −2t − C0 = −2t + C

    |25 − y| = e

    −2t+C = e

    −2t

    e

    C

    y − 25 = ± e

    C e

    −2t

    y = 25 ± e

    C e

    −2t = 25 + Ae−2t

    .

    Here A = ± e

    C = ± e

    −C0

    is some non-zero constant. Since we want y(0) = 40, we

    substitute and solve for A:

    40 = 25 + Ae0

    15 = A,

    and so y = 25+15e

    −2t

    is a solution to the initial value problem. Note that y is never 25, so

    this makes sense for all values of t. However, if we allow A = 0 we get the solution y = 25

    to the differential equation, which would be the solution to the initial value problem if we

    were to require y(0) = 25. Thus, y = 25 + Ae−2t describes all solutions to the differential

    equation ˙y = 2(25 − y), and all solutions to the associated initial value problems.

    Why could we solve this problem? Our solution depended on rewriting the equation

    so that all instances of y were on one side of the equation and all instances of t were on the

    other; of course, in this case the only t was originally hidden, since we didn’t write dy/dt

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