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Integrating Factors and Reduction of Order Math 240 Integrating factors Reduction of order Integrating Factors and Reduction of Order Math 240 — Calculus III Summer 2015, Session II Monday, August 3, 2015 Integrating Factors and Reduction of Order Agenda Math 240 Integrating factors Reduction of order 1. Integrating factors 2. Reduction of order Integrating Factors and Reduction of Order Introduction Math 240 Integrating factors Reduction of order The reduction of order technique, which applies to second-order linear differential equations, allows us to go beyond equations with constant coefficients, provided that we already know one solution. If our differential equation is y 00 + a1 (x)y 0 + a2 (x)y = F (x), and we know the solution, y1 (x), to the associated homogeneous equation, this method will furnish us with another, independent solution. To accomplish the process, we will make use of integrating factors. Integrating Factors and Reduction of Order Integrating factors Math 240 Integrating factors Reduction of order Integrating factors are a technique for solving first-order linear differential equations, that is, equations of the form dy a(x) + b(x)y = r(x). dx Assuming a(x) 6= 0, we can divide by a(x) to put the equation in standard form: dy + p(x)y = q(x). dx The main idea is that the left-hand side looks almost like the result of the product rule for derivatives. If I(x) is another function then d dy dI (Iy) = I + y. dx dx dx The standard form equation is missing an I in front of let’s multiply it by I. dy dx , so Integrating Factors and Reduction of Order Integrating factors Math 240 Integrating factors Reduction of order When we multiply our equation by I, we get dy I + Ip(x)y = Iq(x), dx d (Iy), we need to have so in order for the left-hand side to be dx dI = p(x)I. dx Rearranging this into dI = p(x) dx, I we can solve: R I(x) = c1 e p(x) dx . Since we only need one function I, let’s set c1 = 1. Integrating Factors and Reduction of Order Integrating factors Math 240 Integrating factors Reduction of order Using this I, we rewrite our equation as d (Iy) = q(x)I, dx then integrate and divide by I to get Z 1 y(x) = q(x)I dx + c . I Our I is called an integrating factor because it is something we can multiply by (a factor) that allows us to integrate. Integrating Factors and Reduction of Order Math 240 Example Find a solution to y 0 + xy = xex 2 /2 . Integrating factors Reduction of order 1. Find the integrating factor I(x) = e R x dx = ex 2 /2 . 2. Multiply it into the original equation: d x2 /2 2 2 2 e y = ex /2 y 0 + xex /2 y = xex . dx 3. Integrate both sides: 1 2 y = ex + c. 2 4. Divide by I to find the solution −x2 /2 1 x2 y(x) = e e +c . 2 ex 2 /2 Integrating Factors and Reduction of Order Math 240 Integrating factors Reduction of order Example Solve, for x > 0, the equation xy 0 + 2y = cos x. 1. Write the equation in standard form: 2 cos x y0 + y = . x x 2. An integrating factor is I(x) = e2 ln x = x2 . 3. Multiply by I to get d 2 (x y) = x cos x. dx 4. Integrate and divide by x2 to get x sin x + cos x + c y(x) = . x2 Integrating Factors and Reduction of Order Reduction of order Math 240 Integrating factors Reduction of order We now turn to second-order equations y 00 + a1 (x)y 0 + a2 (x)y = F (x). We know that the general solution to such an equation will look like y(x) = c1 y1 (x) + c2 y2 (x) + yp (x). Suppose that we know y1 (x). We will guess the solution y(x) = u(x)y1 (x). Plugging it into our original equation yields u00 y1 + u0 2y10 + a1 (x)y1 = F (x). If we let w = u0 then we have reduced our second-order equation to the first-order 0 2y1 F (x) 0 w + + a1 w = . y1 y1 Integrating Factors and Reduction of Order Reduction of order Math 240 Integrating factors Reduction of order We may solve 2y10 F (x) w + + a1 w = y1 y1 using the integrating factor technique: 0 I(x) = y12 (x)e and 1 w(x) = I(x) Z x Rx a1 (s) ds I(s)F (s) c1 ds + . y1 (s) I(x) Then integrate w to find u: Z x Z t Z 1 I(s)F (s) u(x) = ds dt + c1 I(t) y1 (s) x 1 ds + c2 . I(s) Integrating Factors and Reduction of Order Reduction of order Math 240 Integrating factors Reduction of order Finally, we get Z x 1 ds + c2 y1 (x) I(s) Z x Z t 1 I(s)F (s) + y1 (x) ds dt. I(t) y1 (s) Using F = 0 gives us the two fundamental solutions Z x 1 y(x) = y1 (x) and y(x) = y1 (x) ds. I(s) And using c1 = c2 = 0, we get a particular solution Z x Z t 1 I(s)F (s) yp (x) = y1 (x) ds dt. I(t) y1 (s) y(x) = u(x)y1 (x) = c1 y1 (x) Integrating Factors and Reduction of Order Math 240 Integrating factors Reduction of order Example Determine the general solution to xy 00 − 2y 0 + (2 − x)y = 0, given that one solution is y1 (x) = x > 0, ex . 1. Set up the equation for w: 2(x − 1) w = 0. w0 + x 2. Solve for w: w(x) = c1 x2 e−2x . 3. Integrate to find Z 1 u(x) = w(x) dx + c2 = − c1 e−2x (1 + 2x + 2x2 ) + c2 . 4 4. Multiply by y1 for the general solution: y(x) = c1 e−x (1 + 2x + 2x2 ) + c2 ex . Integrating Factors and Reduction of Order Math 240 Integrating factors Reduction of order Example Determine the general solution to x2 y 00 + 3xy 0 + y = 4 ln x, x > 0, by first finding solutions to the associated homogeneous equation of the form y(x) = xr . 1. Find y1 (x) = x−1 . 2. Put the equation in standard form by dividing by x2 : y 00 + 3x−1 y 0 + x−2 y = 4x−2 ln x. 3. Set up the equation w0 + x−1 w = 4x−1 ln x. 4. Find w(x) = 4(ln x − 1) + c1 x−1 . 5. Then u(x) = 4x(ln x − 2) + c1 ln x + c2 . 6. Multiply by y1 (x) = x−1 : y(x) = 4(ln x − 2) + c1 x−1 ln x + c2 x−1 .