consider the DE dy/dt = ycos(t)?

Question

Find the solution of the DE that passes through the point (1, 1).

Find the solution of the DE that passes through the point (-1, -1).

Answer 1

same as dy/y = cos t dt
or integrating

ln y = sint + c

or y = e^ sin t * e^c

= k e^ sin t

if it passes through 1 1 then k =e^-1

so y = e^(sin t -1)

it cannot pass through -1 -1

Answer 2

This is a variables separable DE. You rearrange it to
(1/y)dy = (cost)dt
and then integrate each side separately. You only need to put a constant of integration on one side.
You should be able to finish it from there.

Answer 3

dy/dt = ycos(t)
dy/y = cos(t)dt
Ln(y) = sin(t) + c
y = e^(sin(t) + c)
y = ce^sin(t)

1 = ce^sin(1)
1 = ce^0.84147
1 = 2.3198c
c = 0.43108
y ≈ 0.431e^sin(t)

-1 = ce^sin(-1)
-1 = ce^-0.84147
-1 = 0.43108c
c = -2.3198
y ≈ -2.32e^sin(t)

Answer 4

(1/y) dy = cos(t) dt
Integrating,
ln y + c1 = sin(t) + c2, where ln is natural log,
ln y = sin(t) + (c2 - c1) = sin(t) + C
Now, (t,y) = (1,1) i.e 1,1 is a solution,
ln 1 = sin 1 + C = 0
C = -sin 1
Hence, required answer,is. ln y = sin(t) - sin 1
But there is no solution for the next one because, ln (-1) is not defined.