Differential Equation (Seperable and Bernoulli)?

Question

dy/dt = Ay(B − y), where A and B are positive constants and 0 < y < B.
(i) Find the general solution of the logistic equation, treating it as a separable equation. For this I have got to 1/By dy + 1/(B*B -By) dy = A dt, but don't know where to go from there.

(ii)Find the general solution of the logistic equation, treating it as a Bernoulli equation. I get to (dy/dt)(y^-2) - ABy^-1 = -A before I get stuck.

thanks for any help.

Answer 1

i) dy/dt = Ay(B-y)
1/(y(B-y)) dy = A dt
1/(By) dy + 1/(B(B-y)) dy = A dt
1/B ln y - 1/B ln(B-y) = At + C where C is constant

ii) (dy/dt)(y^-2) - ABy^-1 = -A
(dy/dt) (1/y^2) = AB/y - A
(dy/dt) (1/y^2) = (AB - Ay) / y
dy/dt = (AB - Ay) y
1/(y(AB-Ay)) dy = dt
1/(ABy) dy + 1/(B(AB-Ay)) dy = dt
1/(AB) ln y - 1/(AB) ln(AB-Ay) = t + C where C is constant

Answer 2

(i)
As mathsmanretired said the partial fractions are correct.
You get the solution by integrating both sides: LHS with respect to y and RHS with respect to t:
[1/(By) + 1/(B(B-y))]dy = Adt
↔
[1/y +1(B-y) dy = AB dt
→
∫ [1/y +1(B-y)] dy =∫ AB dt
→
ln(y) - ln(B-y) = ABt + C where C is the constant of integration
↔
ln(y/(B-y)) = ABt + C
→
y/(B-y) = C·exp(ABt] where C' = exp(C)
↔
(B-y)/y = C'·exp(-ABt]
↔
y = B/(1+C·exp(-ABt])

(ii)
You can rewrite this equation as
-1/y² · dy/dt + (AB)/y = A
Solve your equation by the substitution
z = 1/y
The derivative of z with respect to t is
dz/dt = dz/dy · dy/dt = - 1/y² · dy/dt
which is the first term in your equation. Substitute 1/y in the 2nd term an get the linear DE:
dz/dt - AB · z = A
Such an DE can be solved by the method of variation of parameters.
In general for a DE of the type:
dy/dt + P(x) y = Q(x)
the solution is (for details follow the 2nd link)
y = exp{-∫ P(x)dx} · [ ∫ Q(x)·exp{∫ P(x)dx} dx + C]
Here
P(x) = AB and Q(x)= A
Hence:
z = exp{-ABt} · [∫ A·exp{ABt}dt + C]
= exp{-ABt} · [1/B·exp{ABt}dt + C]
= 1/B + C·exp{-ABt} with C' = C · B
= (1 + C'·exp{-ABt}) / B
in terms of y
y = 1/z = B/(1+C·exp{-ABt})
which is surprisingly the same as in (i) ;-)

Note:
The general procedure to solve a Bernoulli DE:
dy/dx + P(x) · y = Q(x)·yⁿ
Divide by y^n and substitute z = y ^(1-n)
This leads to the linear DE:
(1-n)· dz/dx + P(x) · z = Q(x)
which can be solved for z as shown above.
Find y = z ^(1/(1-n))

Answer 3

Your separation into partial fractions is correct. The point of doing this is so that each fraction can be integrated on its own. I am sure that you have been taught that int(1/x) = lnx (that is log base e). Each one is an adaption of this. The R.H.S. is just a constant so integrates easily.
Sorry can't help with part 2.

Answer 4

(ii) (dy/dt)(y^-2) - ABy^-1 = -A

put y^ -1 = u

so that -(y^-2) (dy/dt) = du/dt

(y^-2) (dy/dt) = -du/dt

substituting
-du/dt -AB u = -A

or du/dt +AB u = A

solve as a linear equation