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course Mth 279
March 1 around 8:45am.
q_a_06________________________________________
Differential Equations Notes
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Question: `q001. Solve the equation
y ' = sin(t) / 4y, y(0) = 1.
Over what region(s) of the plane does the existence theorem apply, and what is the domain of your solution?
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Your solution:
(dy/dt) = sin(t) / 4y
4 Int(y dy) = Int (sin(t) dt)
4(y^2/2) = -cos(t) + c
y = sqrt (-cos(t)/2 + c)
y(0) = 1
1 = sqrt (-cos(0)/2 + c)
1 = sqrt (c - ½)
1 = c - ½
c = (3/2)
y(t) = sqrt (3/2 - (cos(t)/2))
3/2 - (cos(t)/2) >= 0
Cos(t) <= 3 (no restrictions)
y(t) = sqrt (3/2 - cos(t)/2)
1 <= (3/2 - cos(t)/2) <= 2
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Given Solution:
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Question: `q002. In general air resistance has components which depend on v and on 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?
Over what t interval can we expect this solution to exist?
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Your solution:
Int (dv / v^2+4v-5) = Int (t dt)
Int (dv/(v+5)(v-1)) = (t^2/2) + c
I know you will have to use partial fractions in order to solve this, but I’m having trouble remembering, but I’ll attempt!
Int ((A/(v+5)) + (B/(v-1)) dv)
A(v-1) + B(v+5) = dv + (t^2/2) + c
If v=1, A(0) + B(6) = dv + (t^2/2) + c
If v=-5, A(-6) + B(0) = dv + (t^2/2) + c
The “dv” is what’s throwing me off. You would need to solve for A and B, plug them into the integral and do the integral.
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@& The `dv shouldn't be in the partial fractions. You are integrating f(v) dv, with f(v) = 1 / ( ( v + 5) ( v - 1) ) . This expression needs to be put into the form A / (v + 5) + B / (v - 1), in order to integrate it with respect to v.
If you get rid of the dv you'll get it.
A = 1/6 and B = -1/6.
Then:
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.
NOTE STUDENT SOLUTION TO SIMILAR EQUATION
Separated out
v’/(5+4v+v^2) - t = 0
Int(v’/(5+4v+v^2)), letting u= arctan(v + 2), du = 1/(1 + (v+2)^2) dv
dv = (5+4v+v^2) du, plugging back into our equation
Int((1/(5+4v+v^2))* 5+4v+v^2 du) = Int(1 du) = u = arctan(v + 2)
??????I’m not this is the correct approach but I’ve spent some time on this one and this is the best I can come with???????
v’/(5+4v+v^2) - t = 0, after integrals are taken
arctan(v + 2) - t^2/2 = C
arctan(v+2) = t^2/2 + c
v+2 = tan( t^2/2 + c), finally
v = -2 + tan( t^2/2 + c),solution is
10 = -2 + tan(c)
c = arctan(8) = 1.466
???I don’t believe this is correct, but I can’t see where my error has been?????
@& There is a - sign in front of that 5. See the solution appended below.
However, except that it doesn't answer the question, there's nothing wrong with solving the equation as you have it and comparing to the given solution.
arctan(v + 2) is indeed the antiderivative, easily enough obtained by completing the square to put the integrand into the form 1 / (1 + u^2).
For the integration constant you would get the equation
10 = - 2 + tan(c),
but that leads to c = arcTan(12), which would be about 1.52, not arcTan(8), which as you say is about 1.47 .
The solution would therefore be
y = arcTan(v+2) + arcTan(12), approximately
y = arcTan(v+2) + 1.52.
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Given Solution:
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Question: `q003. Show that the equation
2 y tan(t) dy + (y^2 tan(t) sec (t) + 1 / (2 sqrt(t) ) dt = 0
is of the form M dy + N dt = 0, with M_t = N_y.
Integrate to find the function F(t, y) for which F_t = N, and F_y = M.
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Your solution:
M dy + N dt = 0 if:
N = (y^2 tan(t) sec (t) + 1 / (2 sqrt(t) )
M = 2 y tan(t) dy
N_y = ((y*sin(t)) / (sqrt(t)*(cos(t))^2)
M_t = (2y / (cos(t))^2)
SO NO!
Would I also need to do Int (N dt) and Int (M dy)?!
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This equation is of the given form if M(y) = 2 y tan(t) and N(t) = (y^2 tan(t) sec(t) + 1 / (2 sqrt(t) ) .
M_t = 2 y tan(t) sec(t) and N_y = 2 y tan(t) sec(t)
so the condition holds.
If F_t = N then integral(F_t dt) = integral(N dt) so
F(t) = integral(N dt) = integral( (y^2 tan(t) sec(t) + 1 / (2 sqrt(t) ) dt) = y^2 / cos(t) + sqrt(t) + const.
Since the integral is with respect to t, in which y is treated as a constant, the constant can be any function of y, so
F(t) = y^2 / cos(t) + sqrt(t) + f(y).
Since F_y = M we also get
F(t) = integral(M dy) = integral ( 2 y tan(t) sec(t) dy) = y^2 / cos(t) + g(t), so
F(t) = y^2 / cos(t) + g(t).
We can reconcile the two forms of F(t) by letting f(y) = 0 and g(t) = sqrt(t), so that now
F(t) = y^2 / cos(t) + sqrt(t).
`q 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 / cos(t) + sqrt(t) = c.
We note that this solution is defined as long as t >= 0, so that sqrt(t) is defined, and cos(t) is not zero, which eliminates solutions t = pi/2 + n * pi .
We can solve for y to obtain an explicit solution:
y = sqrt(cos(t) sqrt(t) + c cos(t) ),
where c is an arbitrary constant.
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Question: `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?
What therefore is dy/dt in terms of v and dv/dt?
What is y in terms of v?
Rewrite the original equation in terms of p(t), q(t), v and dv/dt.
Choose m so that the v in the right-hand side has power 0.
Now the equation is linear nonhomogeneous order 1.
`q What integrating factor allows us to solve the equation?
NOTE: This is an example of a Bernoulli Equation, which is of the form
y ' + p(t) y = q(t) y^n.
This equation differs from a first-order linear homogeneous equation by the factor y^n on the right-hand side.
The general method is to change the variable to v = y^m, then choose the value of m that makes the equation linear.
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Your solution:
n = 3 and m is a constant to be determined
(dv/dt) = m*y^(m-1)*(dy/dt)
(dy/dt) = m^(-1)*y^(1-m)*(dv/dt)
=m^(-1)*v^((1-m)/m)*(dv/dt)
(dv/dt) + m*p(t)*v = m*q(t)*v^((m+3-1)/m)
m = (1-3) = (-2)
(dv/dt) - 2*p(t)*v = (-2)*q(t)
y(t) = v(t)^(1/(-2))
I hope I did this right or at least got a good start to it…
@& Very good.
The details in this case:
Using ' for differentiation with respect to t, v = y^m so v ' = (y^m) ' = m y^(m-1) y '.
In terms of dv/dt and dy/dt we have
dv/dt = m y^(m-1) dy/dt.
dv/dt = m y^(m-1) dy/dt so
dy/dt = 1/m * y^(1 - m) * dv/dt.
v = y^m, so y = v^(1/m).
The equation y ' + p(t) y = q(t) y^3 becomes
1/m * y^(1 - m) * dv/dt + p(t) * v^(1/m) = q(t) * v^(3 / m).
This form still includes the y function. Since y = v^(1/m) we get
1/m * v^( (1 - m) / m ) * dv/dt + p(t) * v^(1/m) = q(t) v^(3 / m).
Multiplying through by m v^((m - 1) / m) we get
dv/dt + m p(t) v^(1/m + (m-1)/m) = q(t) v^(3/m + (m - 1) / m), or
dv/dt + m p(t) v = q(t) v^( (m+2) / m).
The power of v on the right-hand side is (m + 2) / m. If m = -2, then the power is 0.
The resulting equation is
dv/dt - 2 p(t) v = q(t).
This equation, which as we have seen is equivalent to the original equation with the substitution v = y^-2, is linear nonhomogeneous order 1.
The coefficient of the linear term (i.e., the v term) is m p(t), so the integrating factor is e^(integral( -2 p(t) ) dt) = e^(-2 integral(p(t) dt).
NOTE: This is an example of a Bernoulli Equation, which is of the form
y ' + p(t) y = q(t) y^n.
This equation differs from a first-order linear homogeneous equation by the factor y^n on the right-hand side.
The general method is to change the variable to v = y^m, then choose the value of m that makes the equation linear.
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Given Solution:
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Question:
`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.
If L = 1000, P(0) = 100 and P(1) = 200, what is the value of k?
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Your solution:
(dP/dt) = kP (L-P)
(dP/(P(L-P)) = k*dt
(A/P) + (B/(L-P)) * dP = k*dt
A(L-P) + B(P) = 1
(AL - AP) + (BP) = 0(P) + 1
(BP) - (AP) = (0P)
B - A = 0 so B will equal A
A * L = 1 so A = (1/L)
[(1/L)/P] + [(1/L)/(L-P)] * dP = k*dt
=ln (P) + ln (L-P) = (L*k*t) + c
@& An antiderivative of 1 / ( L - P ) is - ln | L - P |.
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=ln (P*(L-P)) = L*k + c
=(P*(L-P)) = e^((L*k)*t + c)
=(P*(L-P)) = e^((L*k)*t) + e^(c)
Would e^(c) just be a constant altogether?
I would solve for “P” now, I think?!
@& e^(L k t + c) = e^c * e^(L k t), not e6c + e^(L k t).
You're on the right track though:
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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&#Good work. See my notes and let me know if you have questions. &#