Lecture 10 - Second order linear differential equations

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1 Lecture 10 - Second order linear differential equations In the first part of the course, we studied differential equations of the general form: = f(t, y) In other words, is equal to some expression involving both y and t (example: = y2 + t 2 ). This type of differential equations are called first order differential equations because they provide an expression for the first derivative of y in terms of y and t. Here is a summary of the methods we studied so far: I. First order linear differential equations: = p(t)y + g(t) Form: = ay b Method: substitution rule of integration Form: + p(t)y = g(t) Method: product rule of derivatives II. First order nonlinear differential equations Form: h(y) = g(t) (separable equations) Method: chain rule of derivatives Form: Equations that can be made separable by the substitution u(x) = y x Method: change of variable from y to u Form: Equations that can be made separable by the substitution u(x) = ax + by Method: change of variable from y to u Form: + p(t)y = g(t)yn (Bernoulli equations) Method: change of variable u(x) = y (n 1) In the second part of the course, we will stu methods for solving second order differential equations. A second order differential equation has the general form: 2 = f(t, y, ) In other words, d2 y 2 is equal to some expression involving, y, and t (example: 2 = 5 7y + sin(t)). We will focus mainly on second order linear differential equations because there are very few algorithms available for second order nonlinear differential equations. A second order linear differential equation has the general form: 2 = q(t) + r(t)y + g(t) We will encounter most second order linear differential equations in the form: p(t)y + q(t)y + r(t)y = g(t) If g(t) = 0, then we call such an equation homogeneous. If g(t) 0, we call it nonhomogeneous. (People who are familiar with Linear Algebra, take notice!) We will first stu homogeneous equations where the functions p(t), q(t), and r(t) are constant. ay + by + cy = 0 Problem: Solve the differential equation y y = 0, y(0) = 2, y (0) = 1. Observations: 1. a = 1, b = 1, c = 0 1

2 2 2. We are now given two initial conditions: one for y and another for y. Intuitively, this is because when you take the antiderivative twice, you end up with two constants that need to be solved for. y y = 0 We are looking for a function y that is equal to its second derivative! First guess: y(t) = e t. y (t) = e t y (t) = e t 2. Does y satisfy the initial conditions? No! y(0) = 1 Improved first guess: y(t) = ce t. y (t) = ce t y (t) = ce t 2. Can we find c so that y satisfies the initial conditions? No! Solving for c: y(0) = c = 2 y(t) = 2e t y (t) = 2e t y (0) = 2 Second guess: y(t) = de t. 1. Does it satisfy the differential equation? Yes! y (t) = de t y (t) = de t 2. Can we find d so that y satisfies the initial conditions? No! Solving for d: y(0) = d = 2 y(t) = 2e t y (t) = 2e t y (0) = 2 So far we found two families y 1 (t) = ce t and y 2 (t) = de t (y 1 (t) = ce t is a family of functions because it encompasses infinitely many functions corresponding to different values of c) both of which satisfy the differential equation but neither of which contains a function that satisfies the initial conditions. Here comes the key insight: Add the two families: y(t) = ce t + de t y (t) = ce t de t y (t) = ce t + de t 2. Can we find c and d so that y satisfies the initial conditions? Yes! Solving for c and d: y(0) = c + d = 2 y (0) = c d = 1.

3 3 c + d = 2 and c d = 1 c = 2 d 2 d d = 1 d = 3 2 c = = 1 2 y(t) = 1 2 et e t Fact 1: Suppose v(t) is a solution to p(t)y + q(t)y + r(t)y = 0, then so is w(t) = cv(t) for any constant c. A solution multiplied by a constant is also a solution. Proof: Since v(t) is a solution to p(t)y + q(t)y + r(t)y = 0, we have: We need to verify that p(t)w + q(t)w + r(t)w = 0! p(t)w + q(t)w + r(t)w = p(t)(cv) + q(t)(cv) + r(t)(cv) = p(t)cv + q(t)cv + r(t)cv = c(p(t)v + q(t)v + r(t)v) = c 0 = 0! p(t)v + q(t)v + r(t)v = 0 Fact 2: Suppose v(t) and w(t) are solutions to p(t)y + q(t)y + r(t)y = 0, then so is u(t) = v(t) + w(t). The sum of two solutions is also a solution. Proof: Since v(t) and w(t) are solutions to p(t)y + q(t)y + r(t)y = 0, we have: p(t)v + q(t)v + r(t)v = 0 p(t)w + q(t)w + r(t)w = 0 We need to verify that p(t)u + q(t)u + r(t)u = 0! p(t)u + q(t)u + r(t)u = 0 p(t)(v + w) + q(t)(v + w) + r(t)(v + w) = p(t)(v + w ) + q(t)(v + w ) + r(t)(v + w) = p(t)v + p(t)w + q(t)v + q(t)w + r(t)v + r(t)w = p(t)v + q(t)v + r(t)v + p(t)w + q(t)w + r(t)w = = 0! Fact 3: Suppose y 1 (t) and y 2 (t) are solutions to p(t)y + q(t)y + r(t)y = 0, then so is y(t) = c 1 y 1 (t) + c 2 y 2 (t). A linear combination of two solutions is also a solution! Proof: Combine facts 1 and 2. The general strategy for solving second order linear equations of the form ay + by + cy = 0 will involve, as above, finding two families of functions which satisfy the differential equation and adding them together together to obtain the solution we are looking for. Problem: Solve the differential equation y + 5y + 6y = 0, y(0) = 2, y (0) = 3. y + 5y + 6y = r 2 e rt + 5re rt + 6e rt = (r 2 + 5r + 6)e rt = 0 Therefore the differential equation will be satisfied exactly when r 2 + 5r + 6 = 0! The equation r 2 + 5r + 6 = 0 is called the characteristic equation of the differential equation y + 5y + 6y = 0!

4 4 To obtain r, we solve the characteristic equation: r 2 + 5r + 6 = (r + 2)(r + 3) = 0 So r = 2 and r = 3. Therefore y 1 (t) = c 1 e 2t and y 2 (t) = c 2 e 3t are two families of solutions. y(t) = c 1 e 2t + c 2 e 3t It remains to solve for c 1 and c 2 : y(0) = c 1 + c 2 = 2 y (t) = 2c 1 e 2t 3c 2 e 3t y (0) = 2c 1 3c 2 = 3 c 1 + c 2 = 2 and 2c 1 3c 2 = 3 c 1 = 9 c 2 = 7 y(t) = 9e 2t 7e 3t Problem: Solve the differential equation 4y 8y + 3y = 0, y(0) = 2, y (0) = y 8y + 3y = 4r 2 e rt 8re rt + 3e rt = (4r 2 8r + 3)e rt = 0 The characteristic equation: 4r 2 8r + 3 = 0 r = b ± b 2 4ac 2a = 8 ± = 8 ± 16 8 = 8 ± 8 4 So r = 3 2 and r = 1 2. Therefore y 1 (t) = c 1 e 3 2 t and y 2 (t) = c 2 e 1 2 t are two families of solutions. y(t) = c 1 e 3 2 t + c 2 e 1 2 t It remains to solve for c 1 and c 2 : y(0) = c 1 + c 2 = 2 y (t) = 3 2 c 1e 3 2 t c 2e 1 2 t y (0) = 3 2 c c 2 = 1 2 So c 1 + c 2 = 2 and 3 2 c c 2 = 1 2. c 1 = 1 2 c 2 = 5 2 y(t) = 1 2 e 3 2 t e 1 2 t General strategy for equations ay + by + cy = 0, y(t 0 ) = m, y (t 0 ) = n. (Characteristic equation with two real solutions.) Step 1: Find two families of solutions. ar 2 e rt + bre rt + ce rt = 0 e rt (ar 2 + br + c) = 0 The characteristic equation ar 2 + br + c = 0 has two solutions r 1 r 2. Therefore y 1 (t) = e r1t and y 2 (t) = e r2t are two families of solutions. y(t) = c 1 e r1t + c 2 e r2t Step 2: Solve for c 1 and c 2. y(t 0 ) = c 1 e r1t0 + c 2 e r2t1 = m y (t) = r 1 c 1 e r1t + r 2 c 2 e r2t

5 y (t 0 ) = r 1 c 1 e r1t0 + r 2 c 2 e r2t0 = n (1) c 1 e r1t0 + c 2 e r2t1 = m (2) r 1 c 1 e r1t0 + r 2 c 2 e r2t0 = n We solve for c 1 using equation (1): c 1 = m c 2e r2t0 e r1t0 We substitute into equation (2): r m c 2 e r2t0 1 e r1t0 + r e r1t0 2 c 2 e r2t0 = n r 1 (m c 2 e r2t0 ) + r 2 c 2 e r2t0 = n r 1 m c 2 r 1 e r2t0 + r 2 c 2 e r2t0 = n c 2 e r2t0 (r 2 r 1 ) = n r 1 m c 2 = n r 1m e r2t0 (r 2 r 1 ) The only way we would not be able to obtain a solution for c 2 is if r 2 r 1 = 0. But since r 1 r 2, we have r 2 r 1 0. Thus, we can always find c 1 and c 2 to satisfy the initial conditions! If we can choose the two solutions in such a way that we can satisfy ANY initial conditions, then we call such solutions independent. Two independent solutions serve as building blocks for all other solutions! Thus, once you find two independent solutions, you can build any other solution out of them by choosing the appropriate constants. Two independent solutions are also called a fundamental set of solutions. Questions: 1. What happens if the characteristic equation has one solution or no solutions? 2. From the general strategy, we learned that a differential equation ay + by + cy = 0 with initial conditions y(t 0 ) = m and y (t 0 ) = n always has a solution as long as the characteristic equation has two solutions. Is this solution unique? 5