This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 2
11. In this diﬀerential equation, M (x, y ) = y 2 + 2xy , N (x, y ) = −x2 . Therefore, ∂M = 2y + 2x, ∂y ∂N = −2x, ∂x and so (∂N/∂x − ∂M/∂y )/M = (−4x − 2y )/(y 2 + 2xy ) = −2/y is a function of y . Then µ(y ) = exp − 2 y dy = exp (−2 ln y ) = y −2 . Multiplying the diﬀerential equation by µ(y ) and solving the obtained exact equation, we get y −2 y 2 + 2xy dx − y −2x2 dy = 0 ⇒ ⇒ ⇒ F (x, y ) = −y −2 x2 dy = y −1 x2 + h(x) ∂ ∂F = y −1 x2 + h(x) = 2y −1x + h (x) = y −2 y 2 + 2xy = 1 + 2xy −1 ∂x ∂x h (x) = 1 ⇒ h(x) = x ⇒ F (x, y ) = y −1x2 + x. Since we multiplied given equation by µ(y ) = y −2 (in fact, divided by y 2 ) to get an exact equation, we could lose the solution y ≡ 0, and this, indeed, happened: y ≡ 0 is, clearly, a solution to the original equation. Thus a general solution is y −1x2 + x = c and y ≡ 0. 13. We will multiply the equation by the factor xn y m and try to make it exact. Thus, we have 2xn y m+2 − 6xn+1 y m+1 dx + 3xn+1 y m+1 − 4xn+2 y m dy = 0. We want My (x, y ) = Nx (x, y ). Since My (x, y ) = 2(m + 2)xn y m+1 − 6(m + 1)xn+1 y m , Nx (x, y ) = 3(n + 1)xn y m+1 − 4(n + 2)xn+1 y m , we need 2(m + 2) = 3(n + 1) 76 and 6(m + 1) = 4(n + 2). ...
View
Full
Document
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details