Question: the differential equation for dy/dx= 3y, where y=2 and x= 0
The DE is: dy/dx= 3y
Take the y over to the other side. This gives us,
(1/y) dy/dx = 3
Now integrate both sides wrt x.
int (1/y) (dy/dx).dx = int 3 dx
which becomes,
int (1/y) dy = int 3 dx
Integrating gives us,
ln(y) = 3x + const
It is convient to write the constant value as ln(K), say. Then
ln(y) - ln(K) = 3x
ln(y/K) = 3x, i.e.
y/K = e^(3x), or
y(x) = K.e^(3x)
Initial conditions
at x=0, y = 2.
Substituting for these values into the equation just found,
y(0) = 2 = K.e^0 = K
Therefore K = 2.
The final equation is: y(x) = 2.e^(3x)