Only One Word for Review Review Engineering Differential Equations The Second Test

Euler the Master of Us All

Euler’s Method: Tangent Line Approximation For the initial value problem we begin by approximating solution y =  (t) at initial point t 0. The solution passes through initial point (t 0, y 0 ) with slope f (t 0, y 0 ). The line tangent to solution at initial point is thus The tangent line is a good approximation to solution curve on an interval short enough. Thus if t 1 is close enough to t 0, we can approximate  (t 1 ) by

Euler’s Formula For a point t 2 close to t 1, we approximate  (t 2 ) using the line passing through (t 1, y 1 ) with slope f (t 1, y 1 ): Thus we create a sequence y n of approximations to  (t n ): where f n = f (t n, y n ). For a uniform step size h = t n – t n-1, Euler’s formula becomes

Euler Approximation To graph an Euler approximation, we plot the points (t 0, y 0 ), (t 1, y 1 ),…, (t n, y n ), and then connect these points with line segments. Note (t 0, y 0 ) was the IC

Autonomous Equations and Population Dynamics In this section we examine equations of the form y' = f (y), called autonomous equations, where the independent variable t does not appear explicitly. y’(t) is the gorillaz velocity which depends on gorillaz height y(t). Important heights are the rest points where y’ is zero; when f(y)=0. These are Equilibrium solutions. Example (Exponential Growth): Solution:

Autonomous Equations: Equilibrium Solns Equilibrium solutions of a general first order autonomous equation y' = f (y) are found by locating roots of f (y) = 0. These roots of f (y) are called critical points. For example, the critical points of the logistic equation are y = 0 and y = K. Thus critical points are constant functions (equilibrium solutions) in this setting.

Autonomous Equations: Equilibrium Solns Equilibrium solutions of a general first order autonomous equation y' = f (y) can be found by locating roots of f (y) = 0. These roots of f (y) are called critical points. Phase diagram is the y axis showing where the monkey climbs (f>0), rests (f=0) and falls (f<0) y’ = y(10 – y) Thus critical points are constant functions (equilibrium solutions) in this setting.

Population Models P(t)= fish pop size, b(t) = individ birth rate (births/unit time/fish) d(t) = individ death rate( deaths/unit time/fish) Units for P’(t) are fish/unit time Balance Law

Velocity and Acceleration x(t) height of an object falling in the atmosphere near sea level; time t, velocity v(t) = x’(t), a(t) = x’’(t) accel. Newton’s 2 nd Law: Net F = ma = m(dv/dt)  net force Force of gravity: - mg  downward force Force of air resistance: -  v (opp to v)  upward force Then get eqn for v (F = Force Grav + Resist Force) and x

Velocity and Acceleration We can also get one eqn for x (using F = Force Grav + Resist Force) m x’’ = -mg – ϒ x’ is one second order de for x which is the same as the previous two first order DEs for x and v

Homogeneous Equations, Initial Values Once a solution to a homogeneous equation is found, then it is possible to solve the corresponding nonhomogeneous equation. Thus consider homogeneous linear Diff equations; and in particular, those with constant coefficients (like the previous) Initial conditions typically take the form Thus solution passes through (t 0, y 0 ), and slope of solution at (t 0, y 0 ) is equal to y 0 '.

Characteristic Equation To solve the 2 nd order equation with constant coefficients, we begin by assuming a solution of the form y = e rt. Substituting this into the differential equation, we obtain Simplifying, and hence This last equation is called the characteristic equation of the differential equation. We then solve for r by factoring or using quadratic formula.

General Solution Using the quadratic formula on the characteristic equation we obtain two solutions, r 1 and r 2. There are three possible results: – The roots r 1, r 2 are real and r 1  r 2. – The roots r 1, r 2 are real and r 1 = r 2. – The roots r 1, r 2 are complex. First assume r 1, r 2 are real and r 1  r 2. In this case, the general solution has the form

Theorem Suppose y 1 and y 2 are solutions to the equation and that the Wronskian is not zero at the point t 0 where the initial conditions are assigned. Then there is a choice of constants c 1, c 2 for which y = c 1 y 1 + c 2 y 2 is a solution to the differential equation (1) and initial conditions (2).

Linear Independence and the Wronskian Two functions f and g are linearly dependent if there exist constants c 1 and c 2, not both zero, such that for all t in I. Note that this reduces to determining whether f and g are multiples of each other. If the only solution to this equation is c 1 = c 2 = 0, then f and g are linearly independent.

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

Repeated Roots Recall our 2 nd order linear homogeneous ODE where a, b and c are constants. Assuming an exponential soln leads to characteristic equation: Quadratic formula (or factoring) yields two solutions, r 1 & r 2 : When b 2 – 4ac = 0, r 1 = r 2 = -b/2a, since method only gives one solution:

General Solution When roots are the same, then two linearly independent solutions are e^{rt} and te^{rt} Thus the general solution for repeated roots is

Complex Roots of Characteristic Equation Recall our discussion of the equation where a, b and c are constants. Assuming an exponential soln leads to characteristic equation: Quadratic formula (or factoring) yields two solutions, r 1 & r 2 : If b 2 – 4ac < 0, then complex roots: r 1 = + i , r 2 = - i  Thus

Euler’s Formula; Complex Valued Solutions Substituting it into Taylor series for e t, we obtain Euler’s formula: Generalizing Euler’s formula, we obtain Then Therefore

Real Valued Solutions: The Wronskian Thus we have the following real-valued functions: Checking the Wronskian, we obtain Thus y 3 and y 4 form a fundamental solution set for our ODE, and the general solution can be expressed as

Real Valued Solutions Our two solutions thus far are complex-valued functions: We would prefer to have real-valued solutions, since our differential equation has real coefficients. To achieve this, recall that linear combinations of solutions are themselves solutions: Ignoring constants, we obtain the two solutions

Theorem (Nonhomogenous Des) If Y 1, Y 2 are solutions of nonhomogeneous equation then Y 1 - Y 2 is a solution of the homogeneous equation If y 1, y 2 form a fundamental solution set of homogeneous equation, then there exists constants c 1, c 2 such that

Theorem (General Solution) The general solution of nonhomogeneous equation can be written in the form where y 1, y 2 form a fundamental solution set of homogeneous equation, c 1, c 2 are arbitrary constants and Y is a specific solution to the nonhomogeneous equation.

Thanks Leonhard Euler for his insights and beautiful mathematics. You are number –e^{(pi)i} in my book. Ry Cooder’s music I think it’s going to work (as long as you do) Good luck on Test 2 42