Each video corresponds to two links, one of the form diff_**_** and the other of the form http://youtu.be/*******.
The first probably won't work, as it points to a file on a DVD you might not have.
The second should work, and is played through YouTube.
The equation F_net = - k x translates to the equation x '' = - k/m * x, a second-order linear differential equation.
The equation x ' = t + x is in the form x ' = f(t, x). We don't yet know how to solve this equation.
We can use integration of both sides to solve x ' = t. Our solution includes an integration constant, which allows us to impose and initial condition on he solution. We impose the condition x(0) = 12.
We solve x '' = a , integrating both sides twice. We get two integration constants, one for each integration, and evaluate these constants to impose two initial conditions on our solution. The two conditions are x(0) = 12 and x ' (0) = 7.
F_net = F_frict - c v is translated to the equation x '' = F_frict / m - c / m * x ', and is of the form x '' = f(t, x, x'). While t and x are not explicitly part of this function, the most general form of a second-order equation does include t and x.
If an additional force - k x is present our equation becomes x '' = F_frict / m - c / m * x ' - k x. This still fits into the general form x '' = f(t, x, x '), where both x and t are present.
Another additional force m g cos(omega * t) becomes x '' = F_frict / m - c / m * x ' - k x - g cos(omega t), which does explicitly include the three variables t, x, x ' of the form x '' = f(t, x, x').
We verify that the equation x '' = - k / m * x has general solution x(t) = B cos(sqrt(k/m) * t) + C sin(sqrt(k/m) * t), and note that this solution is equivalent to x(t) = A cos(sqrt(k/m) * t + phi), where A and phi can be expressed (using some basic trigonometric identities) in terms of B, C, and m.
The latter function can easily be graphed, applying shifting and stretching transformations to the basic function x(t) = cos(t).
We show that the equation x ' = x + t cannot be solved by integrating both sides, which would require that the unknown function x be integrated with respect to t.
We impose two conditions on the solution of the equation x '' = - k / m * x, requiring that x(0) = .5 and also that the maximum value of x(t) be 1. We interpret the solution in terms of the circular model of the cosine function, showing how the solution corresponds to oscillatory motion between x = -1 and x = 1, starting at x = .5.
(For the same of simplicity it is not noted in the video that the reference-circle point can start at either of two locations, one resulting in initial motion to the left, the other in initial motion to the right. This well be discussed later in the course when we student second-order equations in more detail).
We apply another set of conditions to the solution of x '' = - k / m * x.
We consider the geometry of the equation x ' = 3 x - 2 t, looking at the behavior of the solution near the point (.8, .4) in the x vs. t plane. We construct in two steps an approximate solution between t = .8 and t = 1.0.
(the first video is blurry for the first 30 second or so; it becomes clear in time to understand the explanation)
We solve the equation x '' = t^2 + 4, imposing two conditions on the resulting general solution.
We also briefly consider possible ways of expressing the form of the equation x '' = -Ffrict / m - c x ', reinforcing the reason x '' = f(t, x, x ') is the most general form of a second-order equation.
We investigate given solutions of four equations of the form y ' + p(t) y = 0.
We consider why we expect a solution to the equation y ' + p(t) y = 0 to be e^(-integral(p(t) dt)). We apply this insight to two equations. We note that a more general solution, including an integration constant, is y(t) = A e^( integral(p(t) dt))
We explain why the equation y ' + p(t) y = 0 is classified as a first-order linear homogeneous differential equation, and identify which of four different equations are and are not of this form.
If we multiply y ' + t y by e^(t^2 / 2) we obtain an expression which is equal to (e^(t^2 / 2) * y) '. The result is that the left-hand side of the equation y ' + t y = g(t) can, if we multiply both sides by e^(t^2 / 2), be put into the form (e^(t^2 / 2) * y) ' = e^(t^2 / 2) * g(t), in which form the left-hand side is easy to integrate.
We apply the idea of the preceding to the expression y ' + cos(t) y, this time multiplying by e^(sin(t) ) and obtaining the easy-to-integrate form (e^(sin(t)) * y) '.
We discuss how e^(sin(t)) is related to the coefficient cos(t) of y.
We consider why we expect a solution to the equation y ' + p(t) y = 0 to be e^(-integral(p(t) dt). We apply this to two equations.
The equation y ' + p(t) y = g(t) can, using the pattern illustrated by the preceding examples, be put into the form (e^(integral(p(t) dt) * y) ' = e^(integral(p(t) dt) ) * g(t). The left-hand side is easily integrated. If the right-hand side can be integrated we obtain a solution to our equation.
This type of equation is first-order linear and nonhomogeneous.
We solve the equation dP/dt = k P to obtain a general solution, and evaluate k and our integration constant to represent the principle function for an initial investment of $1000 at 6% interest (compounded continuously).
This equation is first-order linear and homogeneous.
We solve the equation dP/dt = k P to obtain a solution satisfying P(2) = $800, P(6) = $1100. We investigate the geometry of this equation on the P vs. t coordinate system and compare the geometric solution to the solution of our equation.
The equation dP/dt = k P + m can represent a principle which is earning interest, with additional money also being added.
This equation is first-order linear and nonhomogeneous.
A 10% salt solution flowing into a vat of fixed volume, from which mixed solution overflows, results in a changing concentration of salt in the vat. This results in the first-order differential equation q ' = .10 r - q / V * r.
We consider the geometry of the q ' vs. t function.
If rate of temperature change is proportional to 'temperature difference' we get a nonhomogeneous linear differential equation for temperature vs. clock time. A change of variable changes this equation to a homogeneous equation.
The same clip introduces separable differential equations and their solution by integration.
We investigate the equaion (H ( t, y) ) ' = 0, where y is considered to be an implicit function of t. Some of this clip is blurred and some is slightly cut off at the left edge, during the first part where we are differentiating H(t, y) = t cos(y) + 1 / y. Our equation is put into the form M(t, y) dt + N(t, y) dy = 0.
We follow a process similar to the preceding to express an equation of the form dF = 0, where F(t, y) is a function of y and t. y is considered an implicit function of t. We clearly identify the functions M(t, y) and N(t, y).
We continue the preceding, exploring why M_y = N_t, and use this fact to show how to recover the original F(t, y) function from the equation M(t, y) dt + N(t, y) dy = 0.
An equation of this form, with M_y = N_t, is called an exact equation.
We consider the Bernoulli equation, which is of the form
y ‘ + p(t) y = q(t) * y^n.
The trick to solving this equation is to use the substitution v = y^m. For this substitution we get y = v^(1/m), and v ‘ = m y^(m-1) * y ‘ so that y ‘ = 1/m v ‘ y^(1-m) = 1/m v ‘ v^( (1-m) / m). If we’re careful with the algebra of our exponents we can, by choosing the right value of m, reduce this to a linear equation.
We illustrate three different Bernoulli equations.
We introduce the logistic equation in the form y ‘ = k y ( L – y). This equation is easily solved by separation of variables, using the technique of integration by partial fractions.
Having integrated both sides it is necessary to solve the resulting equation for our function y(t).
We complete the solution for y and begin to examine the behavior of the solution function and how the solution is related to the original conditions of the problem.
We further examine the nature of the solution of the logistic equation.
We have solve the logistic equation in the form y ‘ = k y ( L – y ). We consider other common forms of the logistic equation. P ‘ = r P ( 1 – P / L ), which is the same as for the original model but with a rescaled version of the growth constant.
In previous discussions we have assumed that y < L. We consider solutions for which y > L.
The equation y ‘ = y ( 1 – y ) can be solved either as a logistic equation or a Bernoulli equation.
Using analytic solutions as well as the geometric behavior of the equation, we consider the concept of equilibrium solutions in the context of the logistic equation.
We solve the equation for motion with a constant applied force and a linear drag force (i.e., a drag force which is proportional to the first power of the velocity), which is a first-order linear nonhomogeneousequation in velocity v.
Our solutions approach the equilibrium solution of the form v = F / k. (This value of v is referred to as the ‘terminal velocity’.) Our solution function represents an exponential approach to a terminal velocity.
If the drag force is proportional to the square of velocity the equation is no longer linear, nor is it a Bernoulli equation. However it can be solved by separating variables, using the technique of partial fractions.
(Note that there is one minor error in the following clip, which is corrected in the subsequent clip).
We find the equilibrium solution of our equation, interpreting in terms of terminal velocity.
The solution of our equation can be expressed in terms of terminal velocity and a hyperbolic tangent function.
The velocity of a moving object can be regarded as a function of time, or as a function of position. The differential equation governing the motion of an object can be written with v as a function of time t, or with v as a function of position x. Depending on the nature of the known information one formulation might have advantages over the other.
We consider some examples:
Euler’s Method for numerical approximation of solutions follows in a straightforward manner from the geometry of the equation.
Subsequent links are vacant: