math

posted by .

The particular solution of the differential equation dy/dt equals y over 4 for which y(0) = 20 is

y = 20e−0.25t
y = 19 + e0.25t
y = 20 e0.25t - my answer
y = 20e4t

  • math -

    dy/dt = y/4
    dy/y = 1/44 dt
    lny = t/4 + c
    y = c*e^(t/4)

    y(0)=20, so

    c*e^0 = 10
    c = 20

    y = 20e^(t/4)

    You are correct.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Differential equation

    1.) (2x+y)dy + (2x+y+6)dx=0 2.)(5t+1)t ds + (25t-1)sdt=0 thanks. :)
  2. Differential equation

    variable saparable(separation of variable) 1.) (2x+y)dy + (2x+y+6)dx=0 2.)(5t+1)t ds + (25t-1)sdt=0 I don't the solution of this problem I think it a "repeated expression". Can someone help me on this?
  3. math

    I've worked these problems but I was just wondering if someone could check over them to see if they're correct...if not please help! (u-2)(u+7) my answer: 7u^2-49u (5t-4)^2 my answer: 25t^2-40+16 Thank you :)
  4. Differential Calculus

    A ball thrown up has a height f(t)=250t-25t^2 after t seconds. (1) find the maximum height (2) find the velocity when it hits the ground.
  5. 7th grade math

    25t=25t+5
  6. math

    Find the LCM of (7+5t),(49-25t^2), and (7t-5t)
  7. math

    4b^7/5t^2 x 25t^4/16b Could someone help me with this?
  8. alg 2

    The path that a football takes can be described by the euation h=25t-5t^2 where h is the height in meters, of the football, at t, in seconds. How high is the ball after 3 seconds?
  9. Calculus

    The particular solution of the differential equation dy/dt=y/4 for which y(0) = 20 is y = 20e^-0.25t y = 19 + e^0.25t y = 20 e^0.25t y = 20^e4t
  10. Math

    How high does the ball get after it is hot into the air?

More Similar Questions