Posted by Bree on .
How do you solve the initial value problem?
(dy/dx)2y+4=0, y(1)=4

calculus 
Damon,
I am going to use t instead of x. I can do initial condition better that way
let y = a e^kt + b
dy/dt = a k e^kt
a k e^kt 2 a e^kt 2b + 4 = 0
so b = 2
a k = 2 a
k = 2
y = a e^2t + 2
when t = 1, y = 4
4 = a e^2 +2
a e^2 = 2
a = 2/e^2
so
y = (2/e^2) e^2t + 2
y = 2 e^(2t2) + 2