math
posted by Anonymous .
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

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
Similar Questions

Differential equation
1.) (2x+y)dy + (2x+y+6)dx=0 2.)(5t+1)t ds + (25t1)sdt=0 thanks. :) 
Differential equation
variable saparable(separation of variable) 1.) (2x+y)dy + (2x+y+6)dx=0 2.)(5t+1)t ds + (25t1)sdt=0 I don't the solution of this problem I think it a "repeated expression". Can someone help me on this? 
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! (u2)(u+7) my answer: 7u^249u (5t4)^2 my answer: 25t^240+16 Thank you :) 
Differential Calculus
A ball thrown up has a height f(t)=250t25t^2 after t seconds. (1) find the maximum height (2) find the velocity when it hits the ground. 
7th grade math
25t=25t+5 
math
Find the LCM of (7+5t),(4925t^2), and (7t5t) 
math
4b^7/5t^2 x 25t^4/16b Could someone help me with this? 
alg 2
The path that a football takes can be described by the euation h=25t5t^2 where h is the height in meters, of the football, at t, in seconds. How high is the ball after 3 seconds? 
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 
Math
How high does the ball get after it is hot into the air?