Assignment 21

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11:57:19 Query 11.9.7 (was page 576 #6) x' = x(1-y-x/3), y' = y(1-y/2-x)

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RESPONSE --> Here we find first the nullcline lines where dx/dt and dy/dt are zero by finding where (1-y-x/3) and (1-y/2 - x) are equal to zero. Setting both of these equal to zero and solving for y will give us an equation of a line along which dy/dt and dx/dt are all zero. These are the nullclines.

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12:36:22 describe the phase plane, including nullclines, direction of solution trajectory in each region, and equilibrium points

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RESPONSE --> Having already found the nullclines we then use them to find the equilibrium points, where both are equal to zero. We see that (0,0), (0,2), (3,0), and (0.6, 0.8). This divides the plane into four regions. Starting at the top left region, which I will call region 1, we find the solution trajectory by finding the slopes of each equation in that region by evaluating them at a point inside that region. For region I, the point we will use is(1/4, 3/2) here dx/dt <0 and dy/dt = 0. So dy/dt is horizontal pointing right and dx/dt is is vertical down and the resultant is thus pointing down and to the right. In region II, which is the region above both nullcline, dx/dt <0 and dy/dt <0. This means that dx/dt is horizontal pointing toward the left and dy/dt is vertical pointing down and thus the resultant trajectory is pointing down and toward the origin. In region III which here is below region II, has dx/dt > 0 and dy/dt < 0 so dx/dt is pointing horizontal toward the right and dy/dt is pointing vertically down and the resultant is pointing down and toward the right. In rgion IV which is near the origin below both dx and dy and here dx/dt > 0 and dy/dt >0, so dx/dt is piont horizontal and to the right and dt/dt is vertical and pointing up and then the resultant is pointing up toward the right.

Good. FOr comparison:

dx/dt = x (1 – y – x/3); this expression is 0 when x = 0 or when 1 – y – x/3 = 0. Thus

dx/dt = 0 has 2 solutions: x = 0 and y = -(1/3)x + 1.

When dx/dt = 0 and dy/dt is not 0 (as is the case for both of these solutions) the nullcline has vertical slope.

Therefore, x = 0 (y-axis) is a nullcline with vertical slopes, and

the line segment (for x>=0 and y>=0) y = -(1/3)x + 1 is a nullcline with vertical slopes

dy/dt = 0, which yields horizontal nullclines, similarly has 2 solutions:

dy/dt = 0 for y = 0 and y = -2x+2.

Therefore

y = 0 is a nullcline with vertical slopes, and the line segment (for x>=0 and y>=0) y = -2x + 2 is a nullcline with horizontal trajectories .

The lines x = 0, y = 0, y = -2x + 2 and y = -1/3 x + 1 divide the first quadrant into four regions.

There are 4 equilibrium points where the nullclines (with perpendicular trajectories) intersect:

x = 0 and y = -2x + 2 have perpendicular trajectories and intersect at (0,2) x = 0 and y = 0 have perpendicular trajectories and intersect at (0, 0) y = 0 and y = -1/3 x + 1 have perpendicular trajectories and intersect at (3, 0) y = -2x + 2 and y = -1/3 x + 1 have perpendicular trajectories and intersect at (0.6,0.8)

Using test points in each region, the trajectories are:

Region bounded by (0,1), (0,0), (1,0), and (0.6,0.8): x and y both increasing. Region bounded by (0.6,0.8), (1,0), and (3,0): x increasing and y decreasing. Region bounded by (0,1), (0,2), and (0.6,0.8): x decreasing and y increasing. Region outside (above and to the right of) segments from (0,2) to (0.6,0.8) and (0.6,0.8) to (3,0): x and y both decreasing.

Depending upon where the starting point at t=0 is in each region, all the trajectories will eventually move towards either y approaching 0 and x increasing without limit or towards x approaching 0 and y increasing without limit as t tends towards infinity. In other words for any starting point not at an equilibrium point, either x or y will go towards 0 while the other goes towards infinity.

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12:36:34 describe the trajectories that result

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RESPONSE --> I have done this above. Addendum: I forgot to say that there are also trajectories along the x and y axes that are horizontal right and vertically up respectively.

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12:39:47 Query problem 11.10.22 (3d edition 11.10.19) (was 10.8.10) d^2 x / dt^2 = - g / L * x

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RESPONSE --> I had trouble with this because it seems like both a gravity problem and a spring problem but it resembles the spring style oscillation more so that is the method I used here. So I let w= sqrt(g/l) and then, following the method in the text, had the general solution to be: x(t) = C1 cos (sqrt(g/l) t) + C2 sin (sqrt(g/l) t).

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12:46:41 what is your solution assuming x(0) = 0 and x'(0) = v0?

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RESPONSE --> Using x(0)= 0 made the solution C1 (1) + C2 (0) = 0 and so C1 =0. Then differentating that result gave me: C2 *(sqrt(g/l))*cos (sqrt(g/l) t). Using x'(0)= v0 makes this C2 sqrt(g/l) (1) = v0 and solving for C2 makes this: C2 = v0/sqrt(g/l) and substituting this back in makes the solution : x(t) = (v0/sqrt(g/l)) * sin (sqrt(g/l) t). I'm not completely certain of this result but after trying different things it seems to be the most consistent with the books method.

You've got it, but think in terms of angular frequency omega = sqrt(g / l).

The variable here is t (not, for example, x).

The argument is sqrt(g/L) * t. This is expressed below as just omega * t, with omega = sqrt(g/L).

The equation is thus

x '' (t) = - omega^2 * x(t), where x(t) is a function of t and x '' (t) is the second derivative of that function.

The sine and cosine functions are characterized by second derivatives which are negative multiples of themselves. You can easily verify using the chain rule that the second derivative of B cos(omega t) is - omega^2 B cos(omega t) and the second derivative of C sin(omega t) is - C omega^2 sin(omega t). Thus, any function of the form B cos(omega t) + C sin(omega t) is a solution to the equation x ''(t) = - omega^2 x(t).

The general solution is usually expressed using c1 and c2 as the constants. So let's write it

x(t) = c1 cos(omega * t) + c2 sin(omega * t).

If x(0) = 0 then we have

c1 cos(omega * 0) + c2 sin(omega * 0) = 0 so that c1 * 1 + c2 * 0 = 0, giving us c1 = 0.

The solution for x(0) = 0 is thus

x = c2 sin(omega x) ).

Since v(0) = x ' (0) we have

v0 = c2 * omega cos(omega * 0)

so that c2 = v0 / omega, giving us solution

x = v0 / omega * sin(omega t ).

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12:52:12 What is your solution if the pendulum is released from rest at x = x0?

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RESPONSE --> My best solution to this is to take the solution obtained above and modeling it like a position equation where x(t) is like s(t) in motion equations that meant just like in the latter equation we have s0 as a constant in the equation I here added x0 to the solution equation so that the solution took into account the initial position without affecting the derivatives. This makes: x(t)= v0/(sqrt(g/l)) *sin ((sqrt(g/l) t) + x0.

We obtain the values of c1 and c2 using the given conditions x(0) = x0 and x ' (0) = 0:

Using general solution calculated in previous response:

x(t) = c1*cos[sqrt (g/L)*t] + c2*sin[sqrt (g/L)*t] x(0) = x0 = c1*cos(0) + c2*sin(0) so c1 = x0.

x ' (0) = 0 leads us to the conclusion that c2 = 0.

Thus for the given conditions

x(t) = x0*cos[sqrt (g/l)*t]

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12:59:34 Query problem 11.10.25 (3d edition 11.10.24) (was 10.8.18) LC circuit L = 36 henry, C = 9 farad

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RESPONSE --> First I found the general solution using the values given. This makes: 36 d^2Q/dt^2 + Q/9 = 0 Then proceeded to separate variables and integrate I ended up with: Q(t) = C1 cos ((1/18)t) + C2 sin ((1/18)t).

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13:03:27 what is Q(t) if Q(0) = 0 and *(0) = 2?

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RESPONSE --> Using the solution above and then using the initial values to find the constants C1 and C2. Q(0)= 0 gave: C1 (1) + C2(0) --> C1 =0 . Now I have: C2 sin ((1/18)t). Differentiating to obtain Q'(t) and using the intial values of I(0)= Q'(0)= 2= 1/18 C2 (1) --> C2 = 36. That makes the particular solution: Q(t) = 36 sin ((1/18)t).

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13:13:07 what is Q(t) if Q(0) = 6 and I(0) = 0?

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RESPONSE --> Here I used the general solution and then found the constants from the initial values. So for Q(0)= 6 then made C1 (1) + C2(0) = 0 --> C1 = 6. Differentiating this made I(t)=Q'(t) = C2/18 cos ((1/18)t) - 1/3 sin ((1/18)t) and with Q'(0)=0 C2 = 0. So the particular solution here is: Q(t)= 6 cos ((1/18)t).

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13:13:34 What differential equation did you solve and what was its general solution? And

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RESPONSE --> All of this is given above.

Good. Again think in terms of angular frequency omega.

In LC, RC and RLC circuits Q stands for the charge on the capacitor, R for the resistance of the circuit, L the inductance and C the capacitance.

The voltage due to each element is as follows (derivatives are with respect to clock time):

capacitor voltage = Q / C voltage across resistor = I * R voltage across inductor = I ' * L,

where I is the current. Current is rate of change of charge with respect to clock time, so I = Q '. Note that since I = Q ', we have I ' = Q ''.

This circuit forms a loop, and the condition for any loop is that the sum of the voltages is 0. So the general equation for an RLC circuit is

L Q '' + R Q ' + Q / C = 0.

This equation is solved using the assumption that Q = A e^(k t) for arbitrary constants A and k. This assumption leads to the characteristic equation

L k^2 + R k + 1 / C = 0, a quadratic equation in k which is easily solved.

For an LC circuit, R = 0 and the equation is just

L Q '' + Q / C = 0 and the solutions to the characteristic equation are k = i / sqrt(L C) and k = -i / sqrt(L C), leading to the general solution

Q = c1 cos(omega * t) + c2 sin(omega * t), where omega = sqrt( 1 / (L C) ).

For the given conditions sqrt(1 / (LC) ) = sqrt( 1 / (36 * 9) ) = 1 / 18.

I = Q ‘(t) = -c1 omega sin(omega * t) + c2 omega cos(omega * t). For the given condition I(0) = 0, this implies that c2 = 0 so our function is

Q(t) = -c1 omega sin(omega * t).

The condition Q(0) = 6 gives us

6 = -c1 sin(0) so that c1 = 6 and our function is therefore

Q(t) = 6 sin(omega * t), again with omega = 1/18.

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13:13:45 how did you evaluate your integration constants?

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RESPONSE --> Given above.

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13:18:06 Query problem 11.11.12 (was 10.9.12)general solution of P'' + 2 P' + P = 0

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RESPONSE --> The characteristic equation here is: r^2 + 2r+1=0. Then using r= -1/2b-1/2sqrt(b^2-4c), we obtain the results of r= -1. We only have one r here because the second half of the r solution equation is 0. Then we use the general solution for b^2 - 4c = 0 and end up with: P(t)=(C1 t + C2) e^(-t).

Good. Just for comparison:

Using test solution P = A e^(r t) the equation

P’’ + 2P' + P = 0

yields characteristic equation

r^2 + 2 r + 1 = 0 with repeated solution r = -1.

This gives us a general solution which is a linear combination of the functions e^-t and t e^-t so that

P(t) = c1 * t e^-t + c2 * e^-t.

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13:18:15 what is your general solution and how did you obtain it?

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RESPONSE --> This is given above.

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13:23:35 If not already explained, explain how the assumption that P = e^(rt) yields a quadratic equation.

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RESPONSE --> What happens is when we substitute in Ce^(rt) for P into the equation as a solution we get: r^2Ce^(rt) + b*rCe^(rt) + c*Ce^(rt). Factor out the Ce^(rt) makes this Ce^(rt) (r^2+br+c)=0. Then if C does not = 0 we can divide out the Ce^(rt) leaving just r^2+br+c=0.

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13:25:12 Explain how the solution(s) to to your quadratic equation is(are) used to obtain your general solution.

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RESPONSE --> That is done by creating the characteristic equation and then solving for r using the solution equation given in the book and substituting back in to our general solution.

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13:32:08 Query problem 11.11.36 (was 10.9.30) s'' + 6 s' + c s NO LONGER HERE. USE 11.11.30 s '' + b s ' - 16 s = 0

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RESPONSE --> Creating our characteristic equation which is r^2+br-16 =0. Now we have three possible situations: a) overdamped, b) underdamped, and c) critically damped. So to find the values of b that create those threes situations we start with b^2 - 4(16) = 0 and solve for b: b=+- 8. So if b=+-8 we have situation c: critically damped. If b > 8 or b<-8 we have the situation where b^2 -64 >0 which is situation a: overdamped. If 8>b>-8 then b^2-64 <0 and that makes situation b: underdamped.

The discriminant would be b^2 + 64, not b^2 - 64.

This equation yields characteristic equation

r^2 + b r + - 16 = 0 with solutions r = (-b +- sqrt(b^2 + 64) ) / 2.

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13:32:22 for what values of c is the general solution underdamped?

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RESPONSE --> Given above but for b.

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13:32:53 for what values of c is the general solution overdamped?

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RESPONSE --> Above for b.

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13:33:00 for what values of c is the general solution critically damped?

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RESPONSE --> Above for b.

Since b^2-4c is never = 0, there is no value of b for which the general solution is critically damped.

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13:34:15 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> The applications of differential are very interesting because they are so varied.

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Assignment 21

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Good. FOr comparison:

dx/dt = x (1 – y – x/3); this expression is 0 when x = 0 or when 1 – y – x/3 = 0. Thus

dx/dt = 0 has 2 solutions: x = 0 and y = -(1/3)x + 1.

When dx/dt = 0 and dy/dt is not 0 (as is the case for both of these solutions) the nullcline has vertical slope.

Therefore, x = 0 (y-axis) is a nullcline with vertical slopes, and

the line segment (for x>=0 and y>=0) y = -(1/3)x + 1 is a nullcline with vertical slopes

dy/dt = 0, which yields horizontal nullclines, similarly has 2 solutions:

dy/dt = 0 for y = 0 and y = -2x+2.

Therefore

y = 0 is a nullcline with vertical slopes, and

the line segment (for x>=0 and y>=0) y = -2x + 2 is a nullcline with horizontal trajectories .

The lines x = 0, y = 0, y = -2x + 2 and y = -1/3 x + 1 divide the first quadrant into four regions.

There are 4 equilibrium points where the nullclines (with perpendicular trajectories) intersect:

x = 0 and y = -2x + 2 have perpendicular trajectories and intersect at (0,2)

x = 0 and y = 0 have perpendicular trajectories and intersect at (0, 0)

y = 0 and y = -1/3 x + 1 have perpendicular trajectories and intersect at (3, 0)

y = -2x + 2 and y = -1/3 x + 1 have perpendicular trajectories and intersect at (0.6,0.8)

Using test points in each region, the trajectories are:

Region bounded by (0,1), (0,0), (1,0), and (0.6,0.8): x and y both increasing.

Region bounded by (0.6,0.8), (1,0), and (3,0): x increasing and y decreasing.

Region bounded by (0,1), (0,2), and (0.6,0.8): x decreasing and y increasing.

Region outside (above and to the right of) segments from (0,2) to (0.6,0.8) and (0.6,0.8) to (3,0): x and y both decreasing.

Depending upon where the starting point at t=0 is in each region, all the trajectories will eventually move towards either y approaching 0 and x increasing without limit or towards x approaching 0 and y increasing without limit as t tends towards infinity. In other words for any starting point not at an equilibrium point, either x or y will go towards 0 while the other goes towards infinity.

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You've got it, but think in terms of angular frequency omega = sqrt(g / l).

The variable here is t (not, for example, x).

The argument is sqrt(g/L) * t. This is expressed below as just omega * t, with omega = sqrt(g/L).

The equation is thus

x '' (t) = - omega^2 * x(t), where x(t) is a function of t and x '' (t) is the second derivative of that function.

The sine and cosine functions are characterized by second derivatives which are negative multiples of themselves. You can easily verify using the chain rule that the second derivative of B cos(omega t) is - omega^2 B cos(omega t) and the second derivative of C sin(omega t) is - C omega^2 sin(omega t). Thus, any function of the form B cos(omega t) + C sin(omega t) is a solution to the equation x ''(t) = - omega^2 x(t).

The general solution is usually expressed using c1 and c2 as the constants. So let's write it

x(t) = c1 cos(omega * t) + c2 sin(omega * t).

If x(0) = 0 then we have

c1 cos(omega * 0) + c2 sin(omega * 0) = 0 so that

c1 * 1 + c2 * 0 = 0, giving us

c1 = 0.

The solution for x(0) = 0 is thus

x = c2 sin(omega x) ).

Since v(0) = x ' (0) we have

v0 = c2 * omega cos(omega * 0)

so that c2 = v0 / omega, giving us solution

x = v0 / omega * sin(omega t ).

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We obtain the values of c1 and c2 using the given conditions x(0) = x0 and x ' (0) = 0:

Using general solution calculated in previous response:

x(t) = c1*cos[sqrt (g/L)*t] + c2*sin[sqrt (g/L)*t]

x(0) = x0 = c1*cos(0) + c2*sin(0) so

c1 = x0.

x ' (0) = 0 leads us to the conclusion that c2 = 0.

Thus for the given conditions

x(t) = x0*cos[sqrt (g/l)*t]

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Good. Again think in terms of angular frequency omega.

In LC, RC and RLC circuits Q stands for the charge on the capacitor, R for the resistance of the circuit, L the inductance and C the capacitance.

The voltage due to each element is as follows (derivatives are with respect to clock time):

capacitor voltage = Q / C

voltage across resistor = I * R

voltage across inductor = I ' * L,

where I is the current. Current is rate of change of charge with respect to clock time, so I = Q '. Note that since I = Q ', we have I ' = Q ''.

This circuit forms a loop, and the condition for any loop is that the sum of the voltages is 0. So the general equation for an RLC circuit is

L Q '' + R Q ' + Q / C = 0.

This equation is solved using the assumption that Q = A e^(k t) for arbitrary constants A and k. This assumption leads to the characteristic equation

L k^2 + R k + 1 / C = 0, a quadratic equation in k which is easily solved.

For an LC circuit, R = 0 and the equation is just

L Q '' + Q / C = 0 and the solutions to the characteristic equation are k = i / sqrt(L C) and k = -i / sqrt(L C), leading to the general solution

Q = c1 cos(omega * t) + c2 sin(omega * t), where omega = sqrt( 1 / (L C) ).

For the given conditions sqrt(1 / (LC) ) = sqrt( 1 / (36 * 9) ) = 1 / 18.

I = Q ‘(t) = -c1 omega sin(omega * t) + c2 omega cos(omega * t). For the given condition I(0) = 0, this implies that c2 = 0 so our function is

Q(t) = -c1 omega sin(omega * t).

The condition Q(0) = 6 gives us

6 = -c1 sin(0) so that c1 = 6 and our function is therefore

Q(t) = 6 sin(omega * t), again with omega = 1/18.

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Good. Just for comparison:

Using test solution P = A e^(r t) the equation

P’’ + 2P' + P = 0

yields characteristic equation

r^2 + 2 r + 1 = 0 with repeated solution r = -1.

This gives us a general solution which is a linear combination of the functions e^-t and t e^-t so that

P(t) = c1 * t e^-t + c2 * e^-t.

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The discriminant would be b^2 + 64, not b^2 - 64.

This equation yields characteristic equation

r^2 + b r + - 16 = 0 with solutions

r = (-b +- sqrt(b^2 + 64) ) / 2.

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Since b^2-4c is never = 0, there is no value of b for which the general solution is critically damped.

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&#Good responses. See my notes and let me know if you have questions. &#