Ok, I pretty much have part A and C down (or I have an idea on how to do them) but I might need some help over part B.
Anyways, here is the question with all parts
Let c be a positive number. A differential equation of the form dy/dt = k^(1+c) where k is a positive constant, is called a doomsday equation because the exponent in the expression ky(1+c) is larger than the exponent 1 for natural growth.
(a) Determine the solution that satisfies the initial condition y(0) = y0
(b) Show that there is a finite time t = T (doomsday) such that lim(t->T-)(y(t)) = infinity.
(c) An especially prolific breed of rabbits has the growth term ky^(1.01). If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is the doomsday?
Again, please solve (b) and explain how you did it.
I mean, you could solve all of it as well but that's not really needed. Anyways, thanks for even reading this! :D
~Fg5x3
Anyways, here is the question with all parts
Let c be a positive number. A differential equation of the form dy/dt = k^(1+c) where k is a positive constant, is called a doomsday equation because the exponent in the expression ky(1+c) is larger than the exponent 1 for natural growth.
(a) Determine the solution that satisfies the initial condition y(0) = y0
(b) Show that there is a finite time t = T (doomsday) such that lim(t->T-)(y(t)) = infinity.
(c) An especially prolific breed of rabbits has the growth term ky^(1.01). If 2 such rabbits breed initially and the warren has 16 rabbits after three months, then when is the doomsday?
Again, please solve (b) and explain how you did it.
I mean, you could solve all of it as well but that's not really needed. Anyways, thanks for even reading this! :D
~Fg5x3
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Please verify whether you meant dy/dt = k*y^(1+c)
start with a:
dy/dt = k*y^(1 + c)... separate and integrate
y^-(1+c) dy = k dt ...integrate using the standard rule for x^n
(y^(-c))/(-c) = kt + C... multiply both sides by -c, note that any changes to C don't matter yet since C can be anything
y^-c = -ckt + D... take both sides to the power of -1/c
y = (-ckt + D)^(-1/c)
now we need to solve for D, given that y(0) = y0
y0 = (0 + D)^(-1/c)... take both sides to the -c
D = (y0)^-c
So your answer is:
y = (-ckt + y0^(-c))^(-1/c)
Part b:
note that c is positive, which means the exponent -1/c is negative. Taking zero to an infinite power results in an infinite discontinuity i.e. your "doomsday". So just find the value of t when the base is equal to zero:
-ckt + y0^(-c) = 0
y0^(-c) = ckt
t = y0^(-c)/ck
That t is your T, which is to say your Doomsday
c) in this case c = 0.01. Your Y0=2. Use the fact that there are 16 rabbits after three months (that is, y(3 months) = 16 ) in order to solve for k. Once you have that, you can easily find the doomsday.
start with a:
dy/dt = k*y^(1 + c)... separate and integrate
y^-(1+c) dy = k dt ...integrate using the standard rule for x^n
(y^(-c))/(-c) = kt + C... multiply both sides by -c, note that any changes to C don't matter yet since C can be anything
y^-c = -ckt + D... take both sides to the power of -1/c
y = (-ckt + D)^(-1/c)
now we need to solve for D, given that y(0) = y0
y0 = (0 + D)^(-1/c)... take both sides to the -c
D = (y0)^-c
So your answer is:
y = (-ckt + y0^(-c))^(-1/c)
Part b:
note that c is positive, which means the exponent -1/c is negative. Taking zero to an infinite power results in an infinite discontinuity i.e. your "doomsday". So just find the value of t when the base is equal to zero:
-ckt + y0^(-c) = 0
y0^(-c) = ckt
t = y0^(-c)/ck
That t is your T, which is to say your Doomsday
c) in this case c = 0.01. Your Y0=2. Use the fact that there are 16 rabbits after three months (that is, y(3 months) = 16 ) in order to solve for k. Once you have that, you can easily find the doomsday.