Math 240 Calculus III

Transcription

1 Calculus III Summer 2015, Session II Monday, August 3, 2015

2 Agenda 1. 2.

3 Introduction The reduction of technique, which applies to second- 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 + a 1 (x)y + a 2 (x)y = F (x), and we know the solution, y 1 (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.

4 are a technique for solving first- linear differential equations, that is, equations of the form a(x) dy + b(x)y = r(x). dx Assuming a(x) 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 (Iy) = I dx dx + di dx y. The standard form equation is missing an I in front of dy dx, so let s multiply it by I.

5 When we multiply our equation by I, we get I dy + Ip(x)y = Iq(x), dx so in for the left-hand side to be d dx (Iy), we need to have di dx = p(x)i. Rearranging this into di = p(x) dx, I we can solve: I(x) = c p(x) dx 1 e. Since we only need one function I, let s set c 1 = 1.

6 Using this I, we rewrite our equation as d (Iy) = q(x)i, dx then integrate and divide by I to get y(x) = 1 ( ) 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.

7 Example Find a solution to 1. Find the integrating factor y + xy = xe x2 /2. I(x) = e x dx = e x2 /2. 2. Multiply it into the original equation: d ( ) e x2 /2 y = e x2 /2 y + xe x2 /2 y = xe x2. dx 3. Integrate both sides: e x2 /2 y = 1 2 ex2 + c. 4. Divide by I to find the solution ( ) y(x) = e x2 /2 1 2 ex2 + c.

8 Example Solve, for x > 0, the equation xy + 2y = cos x. 1. Write the equation in standard form: y + 2 x y = cos x x. 2. An integrating factor is I(x) = e 2 ln x = x Multiply by I to get d dx (x2 y) = x cos x. 4. Integrate and divide by x 2 to get x sin x + cos x + c y(x) = x 2.

9 We now turn to second- equations y + a 1 (x)y + a 2 (x)y = F (x). We know that the general solution to such an equation will look like y(x) = c 1 y 1 (x) + c 2 y 2 (x) + y p (x). Suppose that we know y 1 (x). We will guess the solution y(x) = u(x)y 1 (x). Plugging it into our original equation yields u y 1 + u ( 2y 1 + a 1 (x)y 1 ) = F (x). If we let w = u then we have reduced our second- equation to the first- ( ) 2y w a 1 w = F (x). y 1 y 1

10 We may solve ( ) 2y w a 1 w = F (x) y 1 y 1 using the integrating factor technique: and I(x) = y 2 1(x)e x a 1 (s) ds w(x) = 1 x I(s)F (s) ds + c 1 I(x) y 1 (s) I(x). Then integrate w to find u: x 1 t I(s)F (s) u(x) = I(t) y 1 (s) ds dt + c 1 x 1 I(s) ds + c 2.

11 Finally, we get x 1 y(x) = u(x)y 1 (x) = c 1 y 1 (x) I(s) ds + c 2y 1 (x) x 1 t I(s)F (s) + y 1 (x) ds dt. I(t) y 1 (s) Using F = 0 gives us the two fundamental solutions x 1 y(x) = y 1 (x) and y(x) = y 1 (x) I(s) ds. And using c 1 = c 2 = 0, we get a particular solution x 1 t I(s)F (s) y p (x) = y 1 (x) ds dt. I(t) y 1 (s)

12 Example Determine the general solution to xy 2y + (2 x)y = 0, x > 0, given that one solution is y 1 (x) = e x. 1. Set up the equation for w: w 2(x 1) + w = 0. x 2. Solve for w: w(x) = c 1 x 2 e 2x. 3. Integrate to find u(x) = w(x) dx + c 2 = 1 4 c 1e 2x (1 + 2x + 2x 2 ) + c Multiply by y 1 for the general solution: y(x) = c 1 e x (1 + 2x + 2x 2 ) + c 2 e x.

13 Example Determine the general solution to x 2 y + 3xy + y = 4 ln x, x > 0, by first finding solutions to the associated homogeneous equation of the form y(x) = x r. 1. Find y 1 (x) = x Put the equation in standard form by dividing by x 2 : y + 3x 1 y + x 2 y = 4x 2 ln x. 3. Set up the equation w + x 1 w = 4x 1 ln x. 4. Find w(x) = 4(ln x 1) + c 1 x Then u(x) = 4x(ln x 2) + c 1 ln x + c Multiply by y 1 (x) = x 1 : y(x) = 4(ln x 2) + c 1 x 1 ln x + c 2 x 1.