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Transcript
Ch 2.6: Exact Equations & Integrating Factors Consider a first order ODE of the form M ( x, y ) N ( x, y ) y 0 Suppose there is a function such that x ( x, y ) M ( x, y ), y ( x, y ) N ( x, y ) and such that (x,y) = c defines y = (x) implicitly. Then M ( x, y ) N ( x, y ) y dy d x, ( x) x y dx dx and hence the original ODE becomes d x, ( x) 0 dx Thus (x,y) = c defines a solution implicitly. In this case, the ODE is said to be exact. Theorem 2.6.1 Suppose an ODE can be written in the form M ( x, y ) N ( x, y ) y 0 (1) where the functions M, N, My and Nx are all continuous in the rectangular region R: (x, y) (, ) x (, ). Then Eq. (1) is an exact differential equation iff M y ( x, y ) N x ( x, y ), ( x, y ) R (2) That is, there exists a function satisfying the conditions x ( x, y ) M ( x, y ), y ( x, y ) N ( x, y ) (3) iff M and N satisfy Equation (2). Example 1: Exact Equation (1 of 4) Consider the following differential equation. dy x 4y ( x 4 y ) (4 x y ) y 0 dx 4x y Then M ( x, y ) x 4 y , N ( x, y ) 4 x y and hence M y ( x, y ) 4 N x ( x, y ) ODE is exact From Theorem 2.6.1, x ( x, y ) x 4 y , y ( x, y ) 4 x y Thus 1 2 ( x, y ) x ( x, y )dx x 4 y dx x 4 xy C ( y ) 2 Example 1: Solution (2 of 4) We have x ( x, y ) x 4 y , y ( x, y ) 4 x y and 1 2 ( x, y ) x ( x, y )dx x 4 y dx x 4 xy C ( y ) 2 It follows that 1 2 y ( x, y ) 4 x y 4 x C ( y ) C ( y ) y C ( y ) y 2 k Thus ( x, y ) 1 2 1 x 4 xy y 2 k 2 2 By Theorem 2.6.1, the solution is given implicitly by x 2 8xy y 2 c Example 1: Direction Field and Solution Curves (3 of 4) Our differential equation and solutions are given by dy x 4y ( x 4 y ) (4 x y) y 0 x 2 8 xy y 2 c dx 4x y A graph of the direction field for this differential equation, along with several solution curves, is given below. Example 1: Explicit Solution and Graphs (4 of 4) Our solution is defined implicitly by the equation below. x 2 8xy y 2 c In this case, we can solve the equation explicitly for y: y 2 8xy x 2 c 0 y 4 x 17 x 2 c Solution curves for several values of c are given below. Example 3: Non-Exact Equation (1 of 3) Consider the following differential equation. (3xy y 2 ) (2 xy x3 ) y 0 Then M ( x, y) 3xy y 2 , N ( x, y) 2 xy x3 and hence M y ( x, y) 3x 2 y 2 y 3x 2 N x ( x, y) ODE is not exact To show that our differential equation cannot be solved by this method, let us seek a function such that x ( x, y) M 3xy y 2 , y ( x, y) N 2xy x3 Thus ( x, y ) x ( x, y )dx 3xy y 2 dx 3x 2 y / 2 xy2 C ( y ) Example 3: Non-Exact Equation (2 of 3) We seek such that x ( x, y) M 3xy y 2 , y ( x, y) N 2xy x3 and ( x, y ) x ( x, y )dx 3xy y 2 dx 3x 2 y / 2 xy2 C ( y ) Then y ( x, y ) 2 xy x 3 3x 2 / 2 2 xy C ( y ) ? ?? C ( y ) x 3x / 2 C ( y ) x 3 y 3x 2 y / 2 k 3 2 Thus there is no such function . However, if we (incorrectly) proceed as before, we obtain x 3 y xy2 c as our implicitly defined y, which is not a solution of ODE. Integrating Factors It is sometimes possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor (x,y): M ( x, y ) N ( x, y ) y 0 ( x, y ) M ( x, y ) ( x, y ) N ( x, y ) y 0 For this equation to be exact, we need M y N x M y N x M y N x 0 This partial differential equation may be difficult to solve. If is a function of x alone, then y = 0 and hence we solve d M y N x , dx N provided right side is a function of x only. Similarly if is a function of y alone. See text for more details. Example 4: Non-Exact Equation Consider the following non-exact differential equation. (3xy y 2 ) ( x 2 xy) y 0 Seeking an integrating factor, we solve the linear equation d M y N x d ( x) x dx N dx x Multiplying our differential equation by , we obtain the exact equation (3x 2 y xy2 ) ( x3 x 2 y) y 0, which has its solutions given implicitly by x3 y 1 2 2 x y c 2