find the particular solutions of the differential equation:?

Question

y'= 27 - 3y and y(0)=0

Answer 1

First look at the homogeneous version:

y' + 3y = 0

From the characteristic equation, we know that the homogeneous solution will be yh = Ae^(-3x).

Then, to find a solution of y' + 3y = 27, choose a solution that looks like the thing on the other side, for instance, a constant yp = B.

Then yp' = 0, and you have:

0 + 3(B) = 27, so B = 9.

So the combined solution is y = yh + yp = Ae^(3x) + 9. To find A, use your initial condition:

0 = Ae^0 + 9 = A + 9, so A = -9.

Therefore the answer is y = -9e^(-3x) + 9.

Answer 2

y' = 27 - 3y
y = integral (27 - 3y)
y = 27y - 1.5y^2 + C

y(0) = 27(0) - 1.5(0)^2 + C
y(0) = C

Answer 3

y=27y-(3/2)y^2+C
y(0)=0
so C=0
y=27y-(3/2)y^2