Exam Basics 1 There will be 9 questions. 2 The first 3 are on pre-midterm material. 3 The next 1 is a mix of old and new material. 4 The last 5 questions will be on new material since the midterm. 5 60 points total (+3 bonus points)
First Order Differential Equations 1 Linear First Order Differential Equations 1 y + p(t)y = g(t) 2 Let µ(t) = e p(t)dt 3 y=1/µ(t) µ(t)g(t)dt 2 Separable First Order Differential Equations 1 M(x) + N(y)dy/dx = 0 2 N(y)dy = M(x)dx
Exact and Integrating Factor for Exact A differential equation of the form is exact if M(x, y)dx + N(x, y)dy = 0 M y = N x where M y is the partial derivative of the function M(x, y) with respect to y and N x is the partial derivative of N(x, y) with respect to y. If the equation is exact then we can find some function f such that f x = M and f y = N. Then the solution to the differential equation is f = C.
Sometimes an equation is not exact but we can use an integrating factor to make it exact. There are two types of problems you should be able to do. 1 Equations that can be made exact by multiplying by a function µ(x) that only depends on x. This is true only if M y N x N only depends on N. In this case e.g. µ(x) = M y N x µ(x) N y = e 2x + y 1
2 Equations that can be made exact using a given integrating factor. e.g. x 2 y 3 + x(1 + y 2 )y = 0 µ(x, y) = 1/(xy 3 )
Examples of first order differential equations 1 dy/dx = x3 2y x 2 dy/dx = 2x+y 3+3y 2 x 3 dy/dx = 4x3 +1 y(2+3y) 4 For more examples see Pgs 132-133
Autonomous Equations 1 Graph dy/dt as a function of y. Focus on zeros and where the function is positive and negative. 2 Graph the phase line (number line with equilibrium points marked and draw arrows to represent when solutions are increasing or decreasing) 3 Classify the stationary points as stable, unstable or semi-stable. 4 Sketch several solutions based on the increasing and decreasing information from the phase line. Example: dy/dt = y 2 (y 2 1) where < y 0 <
Figure: Plot of dy/dt as a function of y 2 unstable 1 semistable 0 stable -1-2 Figure: Phase Line and sketch of solutions
Figure: Actual Solutions Follow-up Question: What happens to y as t? This depends on the initial condition! Let y(t 0 ) = y 0. For y 0 > 1, y. For 0 < y 0 < 1, y 0. For y 0 < 0, y 1.
Theorems about guaranteed solution intervals First Order Theorem If p and q are continuous on an interval I : α < t < β containing the point t = t 0, then there exists a unique function y that satisfies the IVP y + p(t)y = g(t) for each t in I. y(t 0 ) = y 0 e.g. Find the largest interval in which the IVP has a unique solution. ty + 2y = 4t 2, y(1) = 2
Theorem Let f (t, y) and f / y be continuous for α < t < β, γ < y < δ containing the point (t 0, y 0 ). Then in some interval t 0 h < t < t 0 + h contained in α < t < β, there is a unique solution to the IVP y = f (t, y), where y(t 0 ) = y 0 e.g. Does the theorem guarantee that y = y 1/3 with the initial condition y(0) = 0 has a unique solution? No, since f / y is undefined at y = 0. In fact, y = (2/3t) 3/2 and y = (2/3t) 3/2 are solutions to the IVP.
Theorems for 2nd Order Theorem Consider the IVP y + p(t)y + q(t)y = g(t), y(t 0 ) = y 0, y (t 0 ) = y 0 where p, q, g are continuous on an open interval I containing t 0. Then there exists a unique solution y to this problem and the solution exists throughout I.
Theorems about the Wronskian Theorem Suppose that y 1 and y 2 are two solutions of the differential equation y + p(t)y + q(t)y = 0. Then the family of solutions y = c 1 y 1 (t) + c 2 y 2 )t) with arbitrary coefficients c 1 and c 2 includes every solutions to the above differential equation if and only if there is a point t 0 where the Wronskian of y 1 and y 2 is not zero. If y 1 and y 2 are both solutions to a homogeneous second order differential equation and their Wronskian is nonzero for some point t 0, we call y 1 and y 2 a fundamental solution set.
Just because we can write every solutions as y = c 1 y 1 + c 2 y 2 doesn t mean we can solve every initial value problem. (Why? Because every solution can be expressed via infinitely many initial conditions. Say the solution to the IVP is y(t) = 2t. This satisfies the initial conditions y(0) = 0, y(1) = 2, y(7) = 14,.... Some of these t 0 may have W (y 1, y 2 )(t 0 ) = 0 Theorem Let y + p(t)y + q(t)y = 0, have solutions y 1 and y 2. The the IVP with initial conditions y(t 0 ) = y 0, y (t 0 ) = y 0 has a solutions of the form y = c 1y 1 + c 2 y 2 if and only if W (y 1, y 2 )(t 0 ) 0
Second Order, Linear, Homogeneous Differential Equations with Constant Coefficients ay + by + c = 0 Characteristic Polynomial ar 2 + br + c = 0 Let r 1 and r 2 be the roots of the characteristic polynomial. 1 If r 1 and r 2 are real and distinct, then Let y 1 = e r 1t and y 2 = e r 2t. 2 If r 1, r 2 = λ + µi then y 1 = e λt cosµt and y 1 = e λt sinµt. 3 If r 1 = r 2 then y 1 = e r 1t and y 2 = te r 1t. In each case, y 1 and y 2 form a fundamental solution set so the general solution is y = c 1 y 1 + c 2 y 2.
Examples 1 y 2y + y = 0 2 y y 6y = 0 3 y 10y + 26y = 0
Method of Undetermined Coefficients Given the differential equation y + p(t)y + q(t)y = g(t) we can use the method of undetermined coefficients to find the solutions. First we find the fundamental solution set to the corresponding homogeneous equation. Then we guess the form of the particular solutions. 1 If g(t) = A 0 t n + A 1 t n 1 + + A n 1 t + A n, guess B 0 t n + B 1 t n 1 + + B n 1 t + B n 2 If g(t) = e αt, guess Ae αt 3 If g(t) = sin(αt) or g(t) = cos(αt), guess Asin(αt) + Bcos(αt). If we have a product of the above, guess the product (make sure to have different constants or functions in front of the sin or cos parts.) If we have a sum, break into different problems, finding a g(t) for each term in the sum.
Warning! If the guess is one of the homogeneous solutions, multiply the guess by t. If it is still a solution to the homogeneous solution (in the case of repeated roots), multiply by t again.
What should we guess for the following: 1 3te t (At + B)e t 2 t 2 cos(2t) (At 2 + Bt + C)cos(2t) + (Dt 2 + Et + F )sin(2t) 3 e t (sin(5t)) Ae t (cos(5t)) + Be t (sin(5t)) Once we have our guess, we plug it back in to the differential equation. Then we can solve for the coefficients. Example: y y = t 2 + 3e t
Given y + p(t)y + q(t)y = g(t) we can use variation of parameters to solve. We first find y 1 and y 2 the fundamental solution set. Then the particular is of the form y2 (t)g(t) Y (t) = y 1 W (y 1, y 2 ) dt + y y1 (t)g(t) 2(t) W (y 1, y 2 ) dt As long as we remember our +C s this will in fact be the general solution. Example: Consider t 2 y 2y = 3t 2 1 Show that the corresponding homogeneous equation has fundamental solution set y 1 (t) = t 2, y 2 (t) = t 1. Use variation of parameters to find the general solution.
New stuff since the first midterm 1 Systems of Linear Differential Equations 2 Series Solutions 3 Laplace Transforms
Systems of Linear Differential Equations 1 Know how to find eigenvalues and eigenvectors of a 2x2 matrix. [ ] 2 5 Example: Find eigenvalues and eigenvectors of A = 1 2 2 Know how to find the general solution set to x = Ax in each of the three cases below. Case 1: If A nxn has n-linearly independent eigenvectors. ( ) 1 e.g. A has λ 1 = 1 with v 1 = and λ 2 2 = 4 with ( ) 1 v 2 =. What is the general solution? 4 x = c 1 ( 1 2 ) e t + c 2 ( 1 4 ) e 4t.
Case 2 If A 2x2 has complex conjugates as its eigenvalues. Example: A has λ = 1 ± 2i with eigenvector v = ( 1 + i Let x 1 = 4 Then the general solution is ( 1 ± i 4 ) e 1+2i. Expand into real and imaginary parts. x = c 1 (real part ofx 1 ) + c 2 imaginary part ofx 1 )
Case 3 If A 2x2 has a repeated eigenvalue with only one linearly independent set of eigenvectors, we need to find the generalized eigenvector. [ ] 1 1 Example:Let A =. Then A has eigenvalue 2 with 1 3 ( ) 1 as its only linearly independent eigenvector. Then we 1 have to find the generalized eigenvector by solving (A λi )w = v i.e. ( ) 1 (A 2I )w = 1 The vector w will always have the form w = av + g where a is an arbitrary constant. Then the general solution is x = c 1 ve λt + c 2 (vte λt + ge λt )
Theorems for Systems of First Order Linear Equations Theorem If x (1), x (2),, x (n) are solutions to the system x = P(t)x on α < t < β and W [x (1), x (2),, x (n) ] 0 for some point in that interval, then x (1), x (2),, x (n) form a fundamental set of solutions for that interval.
Series Solution Given a second order linear homogeneous differential equation with polynomial terms P(x)y + Q(x)y + R(x)y = 0 we can use series solution. Assume that the solution has the form y = a n (x x 0 ) n n=0 We need that x 0 is an ordinary point of the differential equation, i.e. P(x 0 ) 0. Then y = y = na n x n 1 n=0 n(n 1)a n x n 2 n=0
Plug this in to the differential equation. Then we re-index as necessary and add. We can then find a recurrence relation for the coefficients. The first two coefficients are free, say a 0 and a 1. If we look at the terms with a 0 we get one solution, the other solution are the terms with a 1. Example: (1 x)y + y = 0 with x 0 = 0 Find the recurrence relation and the first four terms of each of the two solutions.
Theorem If x 0 is an ordinary point of P(x)y + Q(x)y + R(x)y = 0 and Q(x)/P(x) and R(x)/P(x) have a Taylor Series at x 0 then we get a general solution y = a n (x x 0 ) n = a 0 y 1 (x) + a 1 y 2 (x) n=0 where a 0 and a 1 are arbitrary. The radius of convergence for y 1 and y 2 is at least as large as that of Q(x)/P(x) and R(x)/P(x). Example: Find lower bound on radius of convergence of (1 + x 3 )y + 4xy + y = 0 about x 0 = 0 and x 0 = 2.
Euler Equations Euler Equations have the form ay + by + cy = 0 These have x 0 = 0 as a singular point. We guess that the solution y = x r. Plugging this in we get x r (ar 2 + (b a)r + c) = 0 We then find the roots of the quadratic equations.
If the roots are real and distinct r = r 1, r 2 y = c 1 x r 1 + c 2 x r 2 If the roots are repeated r = r 1 y = c 1 x r 1 + c 2 x r 1 ln(x) If the roots are complex r = λ ± µi y = c 1 x λ cos(µ ln x ) + c 2 x λ sin(µ ln x )
Laplace Transform Theorem If f is a piecewise continuous on o t A for all A > 0 and f (t) Ce at for t M then the Laplace transform of f is L{f (t)} = The Laplace transform is linear. 0 e st f (t)dt L{c 1 f 1 + c 2 f 2 } = c 1 L{f 1 } + c 2 L{f 2 }
L{f (t)} = sl{f (t)} f (0) L{f (t)} = s 2 L{f (t)} sf (0) f (0) L{f (n) (t)} = s n L{f (t)} s n 1 f (0) sf (n 2) (0) f (n 1) (0) We can use these facts to solve differential equations. First take Laplace transform of both sides using linearity. Then solve for L{f (t)}. Then use table to find inverse Laplace Transform. You may need to use partial fractions and complete the square. Example: y 2y + 2y = cos t with y(0) = 1 and y (0) = 0
The Unit Step Function (Heaviside Function) For c > 0, we can define the unit stpe function at c. { 0 : t < c u c (t) = 1 : t c Then and L{u c (t)} = e cs s L{u c (t)f (t c)} = e cs L{f } We can use this to solve differential equations with discontinuos forcing functions. E.g. y + 3y + 2y = u 2 (t)
The Dirac Delta Function The dirac δ function is defined to satisfy the following properties: δ(t t 0 ) = 0 t t 0 δ(t t 0 ) = 1 We can use Laplace Transforms to solve differential equations involving δ(t t 0 ) L{δ(t t 0 )} = e st 0 Example: y + 2y + 2y = δ(t π)