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course MTH 279
2/17 1:30 pm
10209 differential equations`q001. Solve the equation
y ' = sin(t) / y, y(0) = 1.
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y dy = sin t dt
Integrate:
1/2 y^2= -cos t+ c
y= sqrt (-2 cos t + c)
y(0)= 1 = sqrt (-2+c) ---> c= 3
y= sqrt (3-2cos t)
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Over what region(s) of the plane does the existence theorem apply?
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The max y can be is sqrt 5 and the min= 1. So R: {(t,y) | -2π ≤ t ≤ 2π, 1≤ y≤ sqrt 5}
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What is the domain of your solution?
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[-2π, 2π]
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`q002. In general air resistance has components which depend on v and to v^2, where v is the velocity of an object. If v ' = -5 t + 4 t v + t v^2, what is the solution of the equation for which v(0) = 10?
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1/(-5 + 4 v + v^2) dv = t dt
1/((v-1)(v+5))
The right side is easily integrated and the left side needs partial fractions to solve, this gives:
1/6 ln((v-1)/(v+5)) = 1/2 t^2+c
((v-1)/(v+5))= Ce^(3t^2)
1-6/(v+5) = Ce^(3t^2)
@& It's not clear how you got from
((v-1)/(v+5))
on the left-had side to
1-6/(v+5)*@
Solve for v:
v= 6/(1-Ce^(3t^2)) - 5
v(0)= 10 = 6/(1-Ce^(3t^2)) - 5 ---> C= 0.6
v= 6/(1-0.6e^(3t^2)) - 5
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Over what t interval can we expect this solution to exist?
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This function is undefined is when 1=0.6e^(3t^2)--> t= 0.4126 since this makes the denominator 0 and after that v becomes neg and approaches -5 as a limit as t-> inf. So it's values actually only make sense or 0≤ t < 0.4126. I'm not sure about my solution after looking at this.
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This equation can be arranged in the form dv / (v^2 + 4 v - 5) = t dt. The denominator can be factored.
It is then possible to use partial fractions to integrate the left-hand side.
The integral on the left is 1/6 ln( (v - 1) / (v+5) ), so the solution is given implicitly by
1/6 ln( (v - 1) / (v+5) ) = t^2 / 2 + c
The solution is
v = (5 e^(c + 3 t^2) + 1) / (1 - e^(c + 3 t^2)),
which can be expressed as
v = (5 A e^(3 t^2) + 1) / (1 - Ae^(3 t^2)),
where A > 0 is equal to e^c.
The exponential function is defined for any real value of its exponent, and 3 t^2 is always real when t is real.
This solution is therefore defined as long as the denominator is nonzero, which is the case as long as A e^(3 t^2) is not equal to 1 (you can verify that this is the case as long as 3 t^2 is not equal to ln(1 / A), so that t is not equal to +- sqrt( -1/3 ln(A); it is worth noting that this cannot be the case if A > 1).
If v(0) = 10, then e^(3 t^2) = 1 so
10 = (A + 1) / (1 - A), with solution A = 9/11.
The solution is therefore
v = (45/11 e^(3 t^2) + 1) / (1 - 9/11 e^(3 t^2)) = (45 e^(3 t^2) + 11) / (11 - 9 e^(3 t^2))
As long as the denominator 11 - 9 e^(3 t^2) is not equal to zero, the solution will exist.
11 - 9 e^(3 t^2) = 0 when e^(3 t^2) = 11/9 so that 3 t^2 = ln(11/9) and t = +-1/3 ln(11/9), or approximately t = +- .067.
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`q003. Show that the equation
2 y tan(t) dy + (y^2 sec^2(t) + 1 / (2 sqrt(t) ) dt = 0
is of the form M dy + N dt = 0, with M_t = N_y.
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M dy= 2 y tan(t) dy
M_t = 2y sec^2 t
N dt= (y^2 sec^2(t) + 1 / (2sqrt(t))
N_y = 2y sec^2 t
Thus M_t= N_y
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Integrate to find the function F(t, y) for which F_t = N, and F_y = M.
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F = y^2 tan y + h(t)
F= y^2 tan y + sqrt t + g(y)
y^2 tan y + h(t)= y^2 tan y + sqrt t + g(y)
h(t) = sqrt t
g(y)= 0
F= y^2 tan y + sqrt t + C
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@& We do get
F(t) = y^2 tan(t) + sqrt(t).
There is more to do:
Our equation is now of the form dF(t, y) = 0, so that its solution is
F(t, y) = c.
This gives us at least an implicit solution.
The implicit solution F(t, y) = c is
y^2 tan(t) + sqrt(t) = c.
We note that this solution is defined as long as t >= 0 is not equal to pi/2, 3 pi / 2 or either of these quantities plus an integer multiple of 2 pi.
We can solve for y to obtain an explicit solution:
y = sqrt(-cotan(t) sqrt(t) + c cotan(t) ),
where c is an arbitrary constant.************
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`q004. Consider the equation y ' + p(t) y = q(t) y^3.
Let v(t) = y(t)^m, where the value of m has yet to be determined.
What is dv/dt?
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dv/dt= m *y^(m-1)*dy/dt
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What therefore is dy/dt in terms of v and dv/dt?
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dy/dt= m^-1*y^(1-m) dv/dt = m^-1*v^((1/m)/m) dv/dt
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What is y in terms of v?
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y= v^(1/m)
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Rewrite the original equation in terms of p(t), q(t), v and dv/dt.
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m^-1*v^((1-m)/m) dv/dt + p(t) v^(1/m) = q(t)v^3
Multiply by m*v^-(-(1-m/m))
dv/dt + p(t)mv^(-(1-m/m))*v^(1/m) = g(t) mv^((-1+m+n)/m)
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Choose m so that the v in the right-hand side has power 0.
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This means:
-1+m+n= 0 --> m= 1-n
n=3 --> m=-2 thus we have:
dv/dt+ mp(t)v = q(t)
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Now the equation is linear non-homogeneous order 1. Solve it accordingly (use an integrating factor, etc.).
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I = e^(m int(p(t) dt))
Multiplying through
(e^(m int(p(t) dt))*v)'= e^(m int(p(t) dt))*m*q(t)
rearranging:
v(t) = (e^(-(m int(p(t) dt)))*Int(e^(m int(p(t) dt))*m*q(t) )
Subbing for y:
y(t)= [(e^(-(m int(p(t) dt)))*Int(e^(m int(p(t) dt))*m*q(t))]^(-1/2)
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`q005. A population function P(t) satisfies dP/dt = k P, as long as space and resources are unlimited. That is, the rate of growth is proportional to the population.
If, however, the population grows too much, it approaches the carrying capacity L of its environment, and its rate of growth becomes also proportional to (L - P). Thus the rate of population growth is jointly proportional to P and (L - P).
This gives us the equation
dP/dt = k P ( L - P ).
This equation is separable. Find its general solution. You will need to use partial fractions.
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After partial fractions and some algebra:
P/(L-P) = e^(L(kt+C) = Ce^(Lkt)
Multiplying and then subtraction L-P
P= LCe^(Lkt)/(1+Ce^(Lkt))
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If L = 1000, P(0) = 100 and P(1) = 200, what is the value of k?
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Subbing in for L:
P= 1000Ce^(1000kt)/(1+Ce^(1000kt))
P(0) = 100 --> C= 1/9
Subbing this in:
P=1000/9e^(1000kt)/(1+1/9e^(1000kt))
P(1)= 200 ---> k = 0.0081
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@& Everything looks good, but my value for k at the end is somewhat different than yours.
Check for comparison:
We get
dP / (P ( L - P) ) = k dt.
We rewrite 1 / (P ( L - P ) ), using partial fractions:
1 / (P ( L - P ) ) = a / P + b / (L - P) for appropriate values of A and B.
a / P + b / (L - P) = ( a ( L - P ) + b P ) / (P ( L - P )) = (a * L + ( b - a ) * P )
1 / (P ( L - P ) ) = (a * L + ( b - a ) * P ) / (P ( L - P ) ).
The denominators of both sides are equal, so their numerators must be equal. So
a * L + (b - a) * P = 1.
This must be so for all values of P. Thus
a * L = 1 and
(b - a) = 0.
We conclude that a = 1 / L and b = - 1 / L.
Thus
1 / (P ( L - P ) ) = (1 / L) / P + (-1 / L ) / (L - P).
Our equation becomes
(1 / L) / P + (-1 / L ) / (L - P) = k dt
Integrating both sides we get
1 / L * ln | P | - 1 /L * ln | L - P | = k t + c
so that
ln | P | - ln | L - P | = k L t + c
ln (| P | / | L - P |) = k L t + c
| P | / | L - P | = e^(k L t + c).
P and L are both non-negative, so as long as P < L we have
P / (L - P) = A e^(k L t), where A > 0 = e^c can be any positive constant.
Solving for P:
P = (L - P) A e^(k L t)
P = L * A e^(k L t) - P * A e^(k L t)
P ( 1 + A e^(k L t)) = L * A e^(k L t)
P = L * A e^(k L t) / (1 + A e^(k L t) ) = L / (A e^(-k L t) + 1)
Note that as t gets large, e^(- k L t) gets small, so that the denominator approaches 1.
The limiting value of P, as t approaches infinity, is thereofore L / 1 = L.
The t = 0 population is L / (A + 1). If we know the initial population P(0) and the carrying capacity L, then, we can find the value of A.
P(0) = L / (A e^(-k L * 0 ) + 1) = L / (A + 1), so
100 = 1000 / (A + 1).
An easy solution gives us A = 9.
Thus our function is
P(t) = 1000 / (9 e^-(1000 k t) + 1).
We also know that P(1) = 200, so
200 = 1000 / (9 e^-(1000 k) + 1).
Thus
200 * (9 e^-(1000 k) + 1) = 1000
e^(-1000 k) = 800 / 1800 = 4/9
-1000 k = ln(4/9)
k = -ln(4/9) / 1000 = .0009 (instructor's very rough mental estimate, not expected to be very accurate).
Our function is therefore
P(t) = 1000 / (9 e^(- .9 t) + 1)
(again based on mental estimates; the exponent -.9 t is in the right ballpark but not expected to be extremely accurate).
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@& You're doing well. Check my notes and see if there's anything you disagree with.*@