the differential ecuation for dy/dx= 3y, where y=2 and x= 0

Question

this is for calculus for biology 2

Answer 1

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)