Question 1.
x*y' = -3*y+4*x^5 is an example of a first-order equidimensional
o.d.e. The standard "trick" for solving equidimensional equations
involves making the substitution x = exp(z), which transforms the
equation into a constant-coefficient o.d.e.
Let x = exp(z)
d/dx = dz/dx * d/dz = exp(-z) d/dz = 1/x * d/dz
or
x*d/dx = d/dz (i.e., x * dy/dx = dy/dz)
Making the appropriate substitutions into your equation yields:
dy/dz + 3*y = 4*exp(5*z)
Let's find the solution to the homogeneous equation first:
dy/dz = -3*y
the variables are easily separated:
dy/y = -3*dz
which integrates to:
ln(y/y0) = -3(z-z0))
or
y = y0*exp(-3(z-z0))
We can incorporate the factor of exp(-3*z0) into the preexponential
factor in the r.h.s., so:
y_h = A*exp(-3*z) is the homogeneous solution, with the constant A to
be determined from the inital condition.
Now let's find the particular solution.
When the forcing term (in this case, exp(5z)) and all its derivatives
form a finite set, we can form a particular solution from a linear
combination of all the members of the set. If a term in the
homogeneous solution appears in this set, that member of the set is
multiplied by the independent variable. In this case, the only member
of the set is exp(5z), so we try as a particular solution:
y_p = B*exp(5z)
substituting this into the diff. eq.:
5*B*exp(5z) = -3*B*exp(5z) + 4*exp(5z)
8*B = 4
B=1/2
The complete solution is then:
y = y_h + p_p = A*exp(-3*z) + 0.5*exp(5*z)
which can also be written as:
y = A*[exp(z)]^(-3) + 0.5*[exp(z)]^5
substituting x = exp(z) to get back the solution in terms of x:
y = A*x^-3 + 0.5x^5
Using the initial condition y(1) = 1 to solve for A:
1 = A*1^-3 + 0.5*1^5
A=0.5
so:
y(x) = 0.5*(x^5 - x^(-3)) is the desired solution.
Question 2.
We are given that "... N'( t) is proportional to the product of the
number of people who have the disease and the number who do not."
Translating this into a diff.eq.:
dN/dt = A*N*(P-N)
where N is the number of people who have the disease, P is the total
population, and A is a contant.
This equation is easily separated:
1/[N*(P-N)] * dN = A*dt
Integrating both sides yields:
1/P*{ln(N/N0) - ln[(P-N)/(P-N0)]} = A*t (taking t0=0)
ln(N) - ln(P-N) - ln(N0) + ln(P-N0) = A*P*t
We are given that P = 19000, and that at t = 0, N0 = 7500. We want to
know how long it will take for N to increase by an additional 7500
people (i.e., for N = 2*N0)
Plugging in N=2*N0, into the above solution, we get for the doubling time:
t = {ln(2*N0) - ln(P-2N0) - ln(N0) + ln(P-N0)}(A*P) , which we can
solve for t once we know A
Use the original diff. eq. to determine A:
800 people/day = A*(N0 people)*(P-N0)people
A = 800/[(N0)*(P-N0)]
Plugging in the values for N0 and P gives the time it will take to
double the number of diseased people as:
t = 9.926 days
In case you are interested, the solution to the differential equation is:
N(t) = P*[N0/(P-N0)]*exp(A*P*t)/{1+[N0/(P-N0)]*exp(A*P*t)}
which can also be written as:
N(t) = P*N0*exp(A*P*t)/{P+N0*[exp(A*P*t)-1]}
(if I did the algebra right!) |
