Consider the following differential equation : dy = -5y

Consider the following differential equation :
dy = -5y ; initial condition; y = 2 at t = 0 The value of y at t = 3 is
dt

dy	= -5y
dt

dy	= -5dt
y

By Integrating above equation, we get
ln y = 5t + c
at t = 0 , y = 2
ln 2 = c
So, ln y = – 5t + ln 2

ln	y	= -5t
	2

	y	= e^-5t
	2

y = 2e^-5t
at t = 3
y = 2 e^-15

	dy	= -5y ; initial condition; y = 2 at t = 0 The value of y at t = 3 is
	dt