SUMMARY of LECTURES

For a particular lecture or a problem session please click on the date of the lecture/problem session.

Lecture on Wednesday, January 29

Equations in which the unknows are functions are called functional equations. Functional equation containing some derivatives of the unknow function, are called differential equations.
Many processes in nature involve changes, hence also rates. Such processes are clearly described by differential equations, which then are refered to as mathematical models of the process.

The mathematical model of a falling object near the surface of the ocean in the precense of atmosphere is given by the equation   mdv/dt=mg-γv  , v(t) being the velocity of the object depending on time t; m - the mass of the object; γ - the drag coefficent. For particular values of the constants entering the equation, the solution is obtained by simple manipulations. As was shown the solution is not unique, and in fact we obtained an infinite one-parametric family of solutions. Graphing the solutions for different values of the parameter (arbitrary constatnt of integration), it is easily noticed that as time evolves all the solution tend to a common assimptote given by the equillibrium solution   v(t)=mg/γ.

The mathematical model of population of mice in the presence of owls, assuming the growth of the population in the absence of predators is proportional to the current population, was found to be desribed by the differential equation   dp/dt=rp-k  . Here p(t) is the population of mice; r is the growth rate, and k is the number of mice killed by owls per unit time. Similar manipulations to those used in the previous model bring to the solutions of this equation.

up

Lecture on Friday, January 31

The direction field for differential equations having the form   dy/dt=ƒ(t,y)  is constructed by drawing short line segments at each point of a rectangular grid in the t,y plane having the slopes equal to the values of the function  ƒ(t,y)  at the corresponding points of the grid. As is observed from the differential equation any solution y(t) passing trough a grid point will have the line segment constracted at that point as it's tangent line. In this sence the direction fields give a first intuitive idea about the behaviour of solutions of the differential equation without actually solving it.

As a general case of the two mathematical models disscused in the previous lecture, the differential equation (i) dy/dt=ay+b  was introduced. By techniques used to solve the differential equation desribing a falling object, the solutions of (i) were obtained, which were expressed by  y=(b/a)+ce^at ; c being the (arbitrary) constant of integration. Specifying an initial condition for the solutions of the differential equation -  y(0)=y_o   (ii) , the constatnt c was determined to be [y_o-(b/a)], thus yielding a unique solution satisfying the equation (i) and the initial condition (ii). Not very suprisingly the solutions to the models introduced in the previous lecture were also obtained from the solutions to (i) with appropriate values of a and b.

For our subsequent study of the subject, several ways of classifying differential equations were discussed. The first classification is based on the number of independent variables the unknown function depends on. In the case of single independent variable the differential equation containes only ordinary derivatives and therefore is called ordinary differential equation. Otherwise the partial derivatives of the unknown function will enter the equation, which is thus called partial differential equation.
Another classification is based on the highest order of the derivatives of the unknown function entering the equation, called the degree of the differential equation.
Finally we distguish between linear and nonlinear equations conditioned by the linear or nonlinear dependence of the function F on the varibales y, y', y"... in the following expression of the differential equation: F(t, y, y', y",..., y⁽ⁿ⁾)=0.

up

Lecture on Monday, February 03

We now turn to a systematic study of differential equations having the form  dy/dt=ƒ(t,y) (i),  which is a first order differential equation solved for the highest derivative. It should be noticed that not all first order differential equations can be solved for the derivative of the unknown function. The equation (i) is called linear, if ƒ(t,y) is linear in the second variable (y-variable). So a general linear first order differential equation will have the following form
dy/dt+p(t)y=g(t)  (ii), where p(t) and g(t) are functions of the independent variable. A special case of the last equation is the equation   dy/dt=ay+b  discussed in the previous lecture, where in general variable coefficents are taken to be constants. A close analysis of the method used to solve this equation showes that it doesnt work in the case of variable coefficents.

However another approach called method of integrating factor can be employed to solve (ii). The idea behind this method is to multiply both sides of the equation by a function μ(t), yet to be determined, so that the equation becomes readily solvable (integrable). After the multiplication the equation looks  μ(t)dy/dt+μ(t)p(t)y=μ(t)g(t). In order to solve it, we try to bring the left hand side of the last equation to the form  {μ(t)y(t)}'  which deduces the following condition on μ(t):  μ(t)dy/dt+μ(t)p(t)y=μ(t)dy/dt+y(t)dμ/dt   ⇒   μ(t)p(t)=dμ/dt , which in turn gives that
  μ(t)=e^∫p(s)ds (*). With this found μ(t) the differential equation has the form  μ(t)y(t)}'=μ(t)g(t)  from which by integrating both sides it's not difficult to solve for y(t), namely   y(t)=(∫μ(s)g(s)ds + c)/μ(t)  (**)   where c is an arbitrary constant. Now given the differential equation (given the functions p(t) and g(t)) one can find the integrating factor μ(t) from (*), and substituting it into (**) find the solution y(t) by caring out the integrations, thus solve the equation (ii).

up

Lecture on Wednesday, February 05

We continue studying the differential equation  dy/dx=ƒ(x,y) (i) This differential equation is called separable if the function ƒ can be represented as  ƒ(x,y)=g(x)h(y), where g(x) is a function only of x and h(y) is a function of only y. For convenience we will study (i) by taking  ƒ(x,y)=g(x)/p(y) (this can be always done by taking p(y)=1/h(y)). Then the differential equation is p(y)dy/dx=g(x)  (ii), which is in a directly integrable form in the following sence. If we denote the antiderivative of p(y) by P(y) (P'=p), and the antiderivative of g(y) by G(y) (G'=g), then the following string of identities obviously holds:
d/dxP(y(x))={P'=p; chain rule}=p(y)dy/dx={equation (ii)}=g(x)={G'=g}=d/dxG(x),
from which it implies that P(y)=G(x)+c as two functions having equal derivatives. Equivalently ∫p(y)dy = ∫g(x)dx + c. This expression is called implicit solution of equation (ii). In some cases it will be possible to solve for y=y(x) which will be the "general" explicit solution of the differential equation.

up

Lecture on Friday, February 07

Differential equations naturally arise in many real world applications. By first analyzing the process involving rates of changes one may translate the physical situation into mathematical terms, thus arriving at the mathematical model of the process. The feasiblity of passing to a mathematical interpretation lies in the relative simplicity of mathematical analysis compared to experimental analysis. It should be noted, however, that mathematical modeling and experiment or observation are both important and in some sence are complimentary in scientific investigation. Generally mathematical modelling consists of the steps illustrated in the following diagram:

While this example has no special significance, it can be viewed as a toy model for more complicated processes in nature, such as the problem of pollutants in the lake. In these real-world applications it's important to bear in mind that the mathematical model gives only approximation to the process, since in this cases the concentration of pollutants in the inflow or the rate may not be easy to determine, as well as not always the sollutions in water mix with a simple pattern (nonuniform).

up

Lecture on Monday, February 10

As another example of applications modeled with first order differential equations we considered the motion of an object with constant mass projectiled from the surface of the earth. Assuming no air resistance but taking into account the change in the grafitational field of earth with distance, we want to find the velocity of the object at any time t as weel as the initial velocity required to lift the body to an altitude ξ, and the smallest initial velocity for which the body will not return to the earth (escape velocity).

Denoting the alltitude of the object by x with the positive direction chosen away from earth, we have the following expression for the grafitational force depending on the distance x from the surface of earth   F(x)= -GmM/(R+x)²   where G is the universal gravitational constant, m, M are the object's and earth's masses correspondingly and R is the radius of earth; the minus sign indicates that the force is directed toward the center of the earth has a negative direction. Since   F(0)= -GmM/R² = -mg at the surface of earth, we can express the constant GM by g arriving at the expression for the gravitational force   F(x)= -mgR²/(R+x)². With the notation v(t) of the velocity of the object at time t, the equation of motion is  mdv/dt = -mgR²/(R+x)². The chain rule then gives dv/dt=dv/dx•dx/dt=vdv/dx. Replacing the derivative with respect to t by the derivative with respect to x in the equation of motion and droping m on the both sides, we get the differential equation desribing the model  vdv/dx = -gR²/(R+x)² with the initial condition v(0)=v_o. The last equation is separable, so by direct integration after separation of variables we arrive at the following implicit solution  v²/2 = gR²/(R+x) + c. Using the initial condition we find

up

Lecture on Wednesday, February 12

So far we've been primarily concerned by solving Initial Value Problems assuming that the solutions exist. In all the examples discussed we found unique solutions satisfying both the differential equation and the initial condition. This raises the question wheter this is true for all initial value problems of first order. We now will give the answer to this question differently for linear and nonlinear equations.

Theorem 1

If the functions p and g are continious on an open interval I:α < t < β containing the point t=to, then there exists a unique function y = φ(t) that satisfies the differential equation y' + p(t)y = g(t) for each t in I, and that also satisfies the initial condition y(to) = yo, where yo is an arbitrary prescribed initial value.

The proof of the theorem is contained in the lecture given on February 03, if we notice that the continuity of p guarantees the existence and differentiability of the integrating factor μ, which along with the continuity of g guarantees the properness of all the integrations, hence proving the existence of the solution. The uniqueness is obvious since we were able to find a unique value of the constant of integration c entering the expression of the solution using the initial condition.

Theorem 2

Let the functions ƒ and ∂ƒ/∂y be continious in some rectangle α < t < β, γ < y < δ containing the point (to,yo). Then, in some interval to - h < t < to + h contained in α < t < β, there is a unique solution y = φ(t) of the initial value problem   y'=ƒ(t,y),    y(to) = yo.

Observe that the hypothesis of Theorem 2 reduce to those in Theorem 1 if the differential equation is linear {ƒ(t,y) = -p(t)y + g(t)}.
These two theorems illustrate some of the differences between linear and nonlinear differential equations. Note that the first theorems states the existence of the solution over the entire interval of continuity of functions entering the equationm while from the second theorem in the nonlinear case the solution may be defined only on some part of the interval of continuity of the functions ƒ and ∂ƒ/∂y.
Another way in which linear equations differ from nonlinear equations is that for first order linear equation it's possible to find a solution contatining an arbitrary constant, from which all posible solutions may be obtained by specifying values for this constant. As we have seen on some examples this is not true in general for nonlinear equations. As we have also seen, for specific nonlinear equations the best we were able to obtain was an implicit solution F(t,y)=0; while in the case of linear differential equations we have an explicit formula for the general solution, from which an explicit solution to any linear differential equation may be obtained.

up

Lecture on Friday, February 14

The differential equation  dy/dt = ƒ(t,y)  is called autonomous if the function ƒ does not depend on the independent variable explicitly, i.e. ƒ(t,y) = ƒ(y). As you notice autonomous equations are separable, hence all the methods for solving separable equations are applicable. We will discuss in detail some autonomous differential equations which arise in population modelling.

The simplest hypothesis concerning rate of change of a population is the proportionality to the current value of population. In this assumption the mathematical model of the population looks  dy/dt = ry  where the coefficent of proportionality r is called the rate of growth or decline, depending on whether it's positive or negative. The solution of this differential equation subject to the initial condition y(0) = y_o is y = y_oe^rt, which predicts an exponential growth of population in the case r > 0. Though under some close-to-ideal conditions this model is quite accurate for short periods of time, it's clear that eventually limitations on space, food supply, etc. will reduce the growth rate.

To take into account the dependence of the growth rate on the population, we replace the constant r in the previous model by a function h(y), and the modified differential equation  dy/dt = h(y)y. Using a linear (simplest) function for h(y) satisfiying the conditions that h(y) behaves like a constant for small y; h(y) decreases as y increases; and h(y) < 0 for large values of y we obtain the following differential equation  dy/dt = (r-ay)y, called logistic equation. It's convenient to rewrite the logistic eqation in an equivalent form  dy/dt = r(1 - y/K)y where K=r/a. The constant r is called the intrinsic growth rate, which is the growth rate in the absence of limiting factors. By setting the right hand side of the equation to 0 and solving for y, we obtain the constant solutions of the equation y = φ₁(t) = 0 and y = φ₂(t) = K which are called equillibrium solutions of the equation, since dy/dt = 0 for this solutions.

up

Lecture on Tuesday, February 18

Snow closing, No lecture or problem session today.

up

Lecture on Wednesday, February 19

We continue to analyze the populaton model described by the differential equation  dy/dt = r(1 - y/K)y, with the population to satisfy the initial condition y(0) = y_o. To get an insight about the behaviour of the solutions we look at the direction field,which in turn is described by the values of the right hand side of the equation  ƒ(y) = r(1 - y/K)y. Obviously ƒ(y) > 0 for 0 < y < K, and ƒ(y) < 0 for y > K, hence the popolation is increasing as long as its value is less than K, and it's decreasing if it's value is greater than K. Furthemore, for the values of y near 0 and K    ƒ(y) ≈ 0, which means that the solution curves are realtively flat, and they become steeper as the value of y leaves the neighborhood of 0 or K.
The next step in investigation of the behaviour of the solutions is the analysis of concavity, which is governed by the sign of the second derivative d²y/dt². DIfferentiating both sides of the equation dy/dt = ƒ(y) with respect to t and using the chain rule to differentiate the right hand side, we get that  d²y/dt² = ƒ'(y)dy/dt = ƒ'(y)ƒ(y), where ƒ(y) = r(1 - y/K)y. Notice that ƒ(y) is a second order polinomial with a negative coefficent of the leading term, so it's graph is a concave down parabola that intersects the y axis at y = 0 and y = K, and the coordinates of its vertex are (K/2, rK/4). Then the function ƒ(y) is positive and increasing for 0 < y < K/2, hence ƒ'(y)ƒ(y) > 0, and the solution curves are concave up;
for K/2 < y < K, ƒ(y) is positive and decreasing, hence ƒ'(y)ƒ(y) < 0, and the solution curves are concave down;
for y > K, ƒ(y) is negative and decreasing so again ƒ'(y)ƒ(y) > 0, and the solution curves are concave up. Recall that Theorem 2 from lecture given on February 12, the fundamental existence and uniqueness theorem, guarantees that two different solutions never intersect, thus a solution that has initial value less than K while increases to the equillibrium solution y = K as t→∞, it never attaines this value for finite time. For this reason the constant K entering the differential equation is called the saturation level or the caring capaciy of the environment.

In many applications it is enough to have the qualitative information about the solution obtained by the methods discussed above, however, if we wish to have a more detailed information about the function of population versur time, we need to solve the differetial equation  dy/dt = r(1 - y/K)y, subject to the initial condition y(0) = y_o. Since thi equation being an autonomous equation is separable, we can solve it by separating variables and integrating: dy/[(1-y/K)y] = rdt, from which the intgration yiellds  ln|y| - ln|1 - y/K| = rt + c, where c is an arbitrary constant. Taking the case y < K we arrive at the implicit solution y/[(1-(y/K)] = Ce^rt, where the constant C = e^c can be obtained from the initial condition giving C = y_o/[1-(y_o/K)]. Using this value of C and solving the implicit solution for y, we obtain the following expression for the solution y = y_oK/[y_o + (K-y_o)e^-rt].
From the last expression for the solution it's not difficult to see that y→K as t→∞,  no matter what the initial value is. For this reason the equillibrium solution y = φ₂(t) = K is called assymtotically stable, while the solution y = φ₁(t) = 0 is called unstable equillibrium solution since the solutions with small initial values depart from this equillibrium solution as time evolves.

up

Lecture on Friday, February 21

We now consider the equation  dy/dt = -r(1 - y/T)y with the initial condition y(0) = y_o , which differs from the equation discussed in last lecture only by the minus sign in front of the intrinsic growth rate r (other than replacing the parameter K by T). As before to analyze the behaviour of solution we look at the graph of ƒ(y) which now is a concave up parabola, since the sign of the leading term is positive. The graph obviously intersects the y axis at the points y = 0 and y = T, and as before these correspond to the constant (equillibrium) solutions of the equation y = ψ₁(t) = 0 and y = ψ₂(t) = T. Furthemore since ƒ(y) < 0 for 0 < y < T, and ƒ(y) > 0 for y > T, the population will decrease if its value is less than T, and will increase otherwise (y > T). For this the parameter T is called a threshold level.
The same concavity analysis as last time yields that the solutions have inflection points as the population reaches the value T/2. Then we can solve the differential equation by the same methods used for the previous equation, but since the equations differ only by the sign of the parameter r, we can get the solutions from those of the previous equation by mere replacement of r by -r (and changing K to T) thus arriving at   y = y_oT/[y_o + (T-y_o)e^rt]. Its is clear from this expression for the solution that the if y_o < T then y→T as t→∞, . Unlike this, the value of a population with initial condition greater than T will become unbounded for some finite time t* which makes the denominator in the expression for the solution equal to zero, from which it can be found to be t* = 1/r ln[y_o/(y_o-T)]. While the populations of some species exhibit the threshold fenomenon in te sence that if too few are present, the species cannot propogate itself, populations cannot become unbounded in finite time. To adjust this dissadventage of this model we combine it with the logistic model, arriving at the equation  dy/dt = -r(1 - y/T)(1 - y/K)y  where 0 < T < K. For this differential equation ƒ(y) is a cubic polynomial and has three critical points: y = 0, y = T and y = K. This model predicts both threshold and environment caring capacity phenomena.

up

Lecture on Monday, February 24

Second order differential equations frequently arise in mathematical physics and engineering. They are also crucial for understanding higher order differential equations, and as will be seen, many of the features of second order differential equations canonicaly generalize to higher orders.
The general second order differential equation has the form F(t, y, y', y") = 0, where F is a function of four variables. An important class of second order differential equations is the class of equations that can be solved for the highest derivative, i.e., can be brought to the form  d²y/dt² = ƒ(t, y, dy/dt)  (i), where ƒ is a function of three variables. We will mostly concern ourselves with the study of the last differential equation, and all the equations under consideration will be assumed to be from this class.
An initial value problem for a second order differential equation will consist of the equation along with a pair of initial conditions y(t_o) = y_o and y'(t_o) = y'_o. Roughly speaking the two conditions are neccesary to determine the values of arbitrary coefficents coming from the two integrations required in soving second order differential equations.
Equation (i) is called linear if the function ƒ has the form ƒ(t, y, dy/dt) = g(t) - p(t)dy/dt - q(t)y, where the functions g, p and q depend only on t; that is, if ƒ is linear in y and y'; the equation is called nonlinear otherwise. So the second order linear differential equation has the following general form

We will now concentrate our attention on the case of constant coefficents, for which the equation (iii') becomes ay" + by' + cy = 0, where a, b and c are constants. After considering some simple equations by assigning special values to the constatnts a, b and c, one comes to the idea of seeking solutions to the equation ay" + by' + cy = 0 in the form y = e^rt, where r is a constant. Substituting this function into the differential equation we arrive at the following condition for the constant r   ar² + br + c = 0, which is a second order algebraic equation called the characteristic equation for the differential equation. Three cases are possible for the roots of this algebraic equation: 1) the roots are real and different; 2) the roots are real but repeated; 3) the roots are complex conjugate.
In the first case we denote the roots by r₁ and r₂ (r₁ ≠ r₂). then y₁ = e^r₁t and y₂ = e^r₂t are two solutions of the equation ay" + by' + cy = 0. From the superposition principle it follows that the function y = c₁e^r₁t + c₂e^r₂t is also a solution for the same differential equation, where the arbitrary constants c₁ and c₂ can be found from the initial conditions y(t_o) = y_o and y'(t_o) = y'_o.

up

Lecture on Wednesday, February 26

We continue to study the second order linear homogeneous differential equations of the form y" + p(t)y' + q(t)y = 0. To simplify notations we introduce the concept of a differential operator. Let the functions p and q be continuous on some open interval I, then for any twice differentiable function φ on I, we define the differential operator L by the equation

Theorem

If the functions p, g and q are continious on an open interval I : α < t < β containing the point t=to, then there is exactly one solution y = φ(t) to the equation y" + p(t)y' + q(t)y = g(t) satisfying the initial conditions y(to) = yo and y'(to) = y'o, and this solution exists troughout the interval I.

As in the case of first order equations, this theorem asserts existence of the solution, its uniqueness, and the fact that the solution is defined on the entire interval where the coefficent functions are continious.
Using the fact that solutions to equations exist, one can derive new solutions to the equations by forming linear combinations of the known solutions. The possibility of forming linear vombinations is given by the superposition principle, which in the operator notation reads:
Superposition Principle:     if L[y₁] = 0 and L[y₂] = 0, then L[c₁y₁ + c₂y₂] = 0 for arbitrary constants c₁ and c₂.

up

Lecture on Friday, February 28

The superposition principle tells us that begining with two solutions one can construct a doubly infinite family of solutions to the equation L[y] = y" + p(t)y' + q(t)y = 0, the natural question then would be if all the solutions to a differential equation are included in this family; i.e. if we have the solutions y₁, y₂ and y do there neccesarily exist constants c₁ and c₂ such that y = c₁y₁ + c₂y₂?
To answer the above question we start by assuming that the solution y satisfying the initial conditions y(t_o) = y_o, y'(t_o) = y'_o can be written in the form y(t) = c₁y₁(t) + c₂y₂(t). Then using the initial conditions one arrives at the following system of equations for the constatnts c₁ and c₂:

W(y1, y2)(to) =

y1(to)  y'1(to)

y2(to) y'2(to)

= y1(to)y'2(to) - y'1(to)y2(to) ≠ 0

Theorem 1

Suppose y1 and y2 are two solutions of the differential equation L[y] = y" + p(t)y' + q(t)y = 0, and that the Wronskian W = y1y'2 - y'1y2 is not zero at the point to where the initial conditions y(to) = yo and y'(to) = y'o are assigned. Then there is a choice of the constants c1, c2 such that the solution y of the equation, satisfying the initial conditions can be expressed in the form y = c1y1(t) + c2y2(t).

By creating "artificial" initial conditions and using this theorem one can establish a more general result, which completelly answers the question stated in the begining of the lecture:

Theorem 2

If y1 and y2 are two solutions of the differential equation L[y] = y" + p(t)y' + q(t)y = 0, and if there is a point to where the Wronskian of y1 and y2 is nonzero, then the family of solutions y = c1y1(t) + c2y2(t) with arbitrary constants c1 and c2, includes every solution of the above equation.

up

Lecture on Monday, March 03

Review for the first midterm exam. The practice exam was also discussed.

up

Lecture on Wednesday, March 05

The notion of fundamental set of solutions is closely related to the concept of linear independence of functions. Two functions ƒ and g are said to be linearly dependent on an interval I if there exist two constants k₁ and k₂, not both zero, such that k₁ƒ(t) + k₂g(t) = 0 for all t in I. The functions ƒ and g are said to be linearly independent on an interval I, if k₁ƒ(t) + k₂g(t) = 0 for all t in I implies that k₁ = k₂ = 0; that is, the are not linearly dependent.
This definitions generelazie to any number of functions in a straightforward way, though in the case of two functions it's easy to see that they are linearly dependent on some interval if they are proportional to each other on that interval, and independent otherwise.
The connection between linear independence of two differentiable functions and their Wronskian is given by the following

Theorem 1

If ƒ and g are differentiable functions on some open interval I and if W(ƒ, g)(to) ≠ 0 for some point to in I, then ƒ and g are linearly independent on I. Moreover, if ƒ and g are linearly dependent on I, then W(ƒ, g)(to) = 0 for all t in I.

Since we are interested in the question of whether two solutions to a second order linear homogeneous differential equations form a fundamental set, or equivalently, whether their Wronskian is nonzero, we will explore some more useful properties of W(y₁, y₂)(t).

Theorem 2

(Abel's Theorem) If y1 and y2 are solutions of the differential equation  L[y] = y" + p(t)y' + q(t)y = 0, where p and q are continious on an open interval I, then the Wronskian W(y1, y2)(t) is given by W(y1, y2)(t) = c exp [ -∫ p(t)dt ]. where c is a certain constant that depends on y1 and y2, but not on t. Furthemore, W(y1, y2)(t) is either zero for all t in I (if c = 0) or else is never zero in I (if c ≠ 0).

The fact that y₁ and y₂ are solutions of the equation, along with some simple manipulations brings to the following condition for the Wronskian,   W' + p(t)W = 0, which is a first order differential equation with the unknown function being W(y₁, y₂)(t). Solving this differential equation one arrives exactly at the form of the Wronskian stated in the theorem.

Combining the above two theorems with some extra work using the existence and uniqueness theorem one can proof

Theorem 3

Let y1 and y2 be solutions of the differential equation  L[y] = y" + p(t)y' + q(t)y = 0, where p and q are continious on an open interval I. Then y1 and y2 are linearly dependent on I if and only if W(y1, y2)(t) is zero for all t in I. Alternatively, y1 and y2 are linearly independent on I if and only if W(y1, y2)(t) is never zero in I.

And we summarize the facts about fundamental sets of solutions, linear independence and Wronskians as follows. Let y₁ and y₂ be solutions of the differential equation  L[y] = y" + p(t)y' + q(t)y = 0, where p and q are continious on an open interval I, then the following four statements are equivalent:

1. the functions y₁ and y₂ form a fundamental set of solutions on I.
2. the functions y₁ and y₂ are linearly independent on I.
3. W(y₁, y₂)(t_o) ≠ 0 for some t_o in I.
4. W(y₁, y₂)(t) ≠ 0 for all in I.

up

Lecture on Friday, March 07

In the lecture on February 24 it was shown how to solve (find the set of fundamental solutions of) the equation ay" + by' + cy = o when the roots of the caracteristic equation ar² + br + c = 0 are real and different. Now we consider the case when the two roots r₁ and r₂ of the caracteristic equation are equal. This happens when the discriminant of the quadratic equation D = b² - 4ac is zero, and then by the quadratic formulas one has that r₁ = r₂ = -b/2a. It is apparent that both of this roots will yield the same (i.e. only one) solution   y₁ = e^-bt/2a. To form the general solution of the equation, another linearly independent from y₁ solution must be found, which we will now do by a method due to d`Alambert.
First notice that by the supperposition principle cy₁ is also a solution to the differential equation for any arbitrary constant c. Then we seek a solution to the differential equation by varying the arbitrary constant c; that is, we look for a solution having the form

W(y1, y2)(t) =

e-bt/2a (-b/2a)e-bt/2a

te-bt/2a (1 - bt/2a)e-bt/2a

= e-bt/a

It should be noted that the procedure used above for equations with constant coefficents is more generally applicable. Suppose we know one solution y₁(t) of the equation y" + p(t)y' + q(t)y = 0. To find the second solution we let   y = v(t)y₁(t). Then

up

Lecture on Monday, March 10

Having developed some theory for solving linear homogeneous second order differential equations, we now return to the nonhomogeneous equations

Theorem 1

If Y1 and Y2 are two solutions of the nonhomogeneous equation (i), then their difference Y1 - Y2 is a solution of the corresponding homogeneous equation. If in addition y1 and y2 are a fundamental set of solutions of the homogeneous equation, then there is a choice of constants c1 and c2, such that Y1(t) - Y2(t) = c1y1(t) + c2y2(t),     for all t in I.

The proof of the Theorem relies only on the linearity of the operator L[y], since Y₁ and Y₂ solve (i), we have that L[Y₁] = g(t) and L[Y₂] = g(t). Hence L[Y₁ - Y₂] = L[Y₁] - L[Y₂] = g(t) - g(t) = 0, which implies that Y₁ - Y₂ is a solution of the homogeneous differential equation. And finally Y₁ - Y₂, being a solution of a homogeneous equation, can be expressed as a linear combination of any two solutions forming a fundamental set, which proves the posibility of the choice of constatnts.

Using the previous result one can completelly charachterize the solutions of the nonhomogeneous differential equations, using a particular solution of the nonhomogeneous equation, and those of the corresponding homogeneous equation. This fact is compactly given by the following

Theorem 2

The general solution of the nonhomogeneous equation (i) can be written in the form y = φ(t) = c1y1(t) + c2y2(t) + Y(t), where y1 and y2 are a fundamental set of solutions of the corresponding homogeneous equation (ii), c1 and c2 are arbitrary constants, and Y is some specific solution of the nonhomogeneous equation (i).

To prove the Theorem note that identifying Y₁ in the previous theorem by φ(t), and Y₂ by the specific solution Y, one arrives at exactly the needed expression

The last theorem states that to find the general solution of the nonhomogeneous equation we must perform three steps:

1. Find the general solutions c₁y₁(t) + c₂y₂(t) of the corresponding homogeneous equation. This solution is sometimes called the complementary solution and is denoted by y_c(t).
2. Find the single solution Y(t) of the nonhomogeneous eqaution. This solution is sometimes reffered to as the particular soluton and is denoted by y_p(t).
3. Add together the functions found in the two preceding steps to find the general solution y_c(t) + y_p(t).

up

Lecture on Wednesday, March 12

We now study a method for finding particular solutions of the nonhomogeneous differential equation

Theorem

If g(t) = g1(t) + g(t), and Y1 and Y2 are particular solutions of the equations ay" + by' + cy = g1(t) and ay" + by' + cy = g2(t) respectively. Then Y1 + Y2 is a particular solution of the equation ay" + by' + cy = g(t)

The proof is obvious upon substitution of Y₁ + Y₂ into the equation ay" + by' + cy = g(t). It also clear that analogous result holds for any arbitrary finite number of summands g(t) = g₁(t) + g₂(t) + ... + g_n(t).

We now specify the Method of Undetermined Coefficents, which can be broken into several steps:

1. Make sure that the function g(t) is from the class of functions which involve nothing more than exponential functions, sines, cosines, polyomials, or sums or products of such functions.
2. If g(t) = g₁(t) + g₂(t) + ... + g_n(t), then form n subproblems, each of which containes only one of the terms g₁(t), ..., g_n(t). The ith problem consisting of the equation ay" + by' + cy = g_i(t) where i runs from 1 to n.
3. For the ith subproblem assume that the particular solution Y_i(t) consists of the appropriate exponential function, sine, cosine, polynomial, or combination thereof given by the following table:
   

   gi(t)		Yi(t)
   Pn(t) = aotn + a1tn-1 + ... + an		ts(Aotn + A1tn-1 + ... + An)
   Pn(t)eαt		ts(Aotn + A1tn-1 + ... + An)eαt
   Pn(t)eαtsinβt Pn(t)eαtcosβt		ts[(Aotn + A1tn-1 + ... + An)eαtcosβt + (Botn + B1tn-1 + ... + Bn)eαtsinβt]

   where the exponent s can be 0,1 or 2, and the factor of t^s is to take care of the case when Y_i(t) in the assumed form duplicates a solution of the homogeneous equation (in which case the substitution of the expression doesnt give any conditions for finding the unspecified coefficents), and is supposed to remove this duplication.
4. Find a particular solution Y_i(t) for each of the subproblems (determine the unspecified coefficents). Then the sum Y₁ + ... + Y_n is a particular solution of the nonhomogeneous equation (i).

up

Lecture on Friday, March 14

The class was a review of the material needed for the assigned homework.
Enjoy the Spring Break!

up

Lecture on Monday, March 24

We continue exploring the ways of solving the nonhomogeneous differential equation

Theorem

If the functions p, q and g are continious on some open interval I, and the functions y1 and y2 form a fundamental set of sollutions of the homogeneous differential equation corresponding to the eqaution y" + p(t)y' + q(t)y = g(t)       (i) then a particular solution of (i) is Y(t) = -y1∫y2g(t)/W(y1, y2)(t)dt + y2∫y1g(t)/W(y1, y2)(t)dt and the general solution is y = c1y1(t) + c2y2(t) + Y(t).

up

Lecture on Wednesday, March 26

Today we come back to the linear homogeneous differential equation of second order ay" + by' + cy = o. As we've already seen, the question of solving this equation reduces to the question of finding the root s of the characteristic equation ar² + br + c = 0. The cases when the roots are real and distinct, or real repeated were completelly resolved in earlier lectures. Now we study the case when the roots are complex conjugate:

up