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

Question

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

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

1. 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