Suppose \u03bc does not depend on x Then \u03bc x 0 Therefore \u03bc N x M y \u03bc y M Using the

Suppose μ does not depend on x then μ x 0 therefore

  • Harvard University
  • ECON 232
  • eeasysemester
  • 70
  • 100% (9) 9 out of 9 people found this document helpful

This preview shows page 56 - 59 out of 70 pages.

Suppose μ does not depend on x . Then μ x = 0. Therefore, μ ( N x - M y ) = μ y M . Using the assumption that ( N x - M y ) /M = Q ( y ), we can find an integrating factor μ by choosing μ which satisfies μ y = Q . We conclude that μ ( y ) = exp R Q ( y ) dy is an integrating factor of the differential equation. 24. Suppose μ is an integrating factor which will make the equation exact. Then multiplying the equation by μ , we have μMdx + μNdy = 0. Then we need ( μM ) y = ( μN ) x . That is, we need μ y M + μM y = μ x N + μN x . Then we rewrite the equation as μ ( N x - M y ) = μ y M - μ x N . By the given assumption, we need μ to satisfy μR ( xM - yN ) = μ y M - μ x N . This equation is satisfied if μ y = ( μx ) R and μ x = ( μy ) R . Consider μ = μ ( xy ). Then μ x = μ 0 y and μ y = μ 0 x
Image of page 56
2.6. EXACT EQUATIONS AND INTEGRATING FACTORS 73 where 0 = d/dz for z = xy . Therefore, we need to choose μ to satisfy μ 0 = μR . This equation is separable with solution μ = exp( R R ( z ) dz ). 25.(a) Since ( M y - N x ) /N = 3 is a function of x only, we know that μ = e 3 x is an integrating factor for this equation. Multiplying the equation by μ , we have e 3 x (3 x 2 y + 2 xy + y 3 ) dx + e 3 x ( x 2 + y 2 ) dy = 0 . Then M y = e 3 x (3 x 2 + 2 x + 3 y 2 ) = N x . Therefore, this new equation is exact. Integrating M with respect to x , we conclude that ψ = ( x 2 y + y 3 / 3) e 3 x + h ( y ). Then ψ y = ( x 2 + y 2 ) e 3 x + h 0 ( y ) = N = e 3 x ( x 2 + y 2 ). Therefore, h 0 ( y ) = 0 and we conclude that the solution is given implicitly by (3 x 2 y + y 3 ) e 3 x = c . (b) 26.(a) Since ( M y - N x ) /N = - 1 is a function of x only, we know that μ = e - x is an integrating factor for this equation. Multiplying the equation by μ , we have ( e - x - e x - ye - x ) dx + e - x dy = 0 . Then M y = - e - x = N x . Therefore, this new equation is exact. Integrating M with respect to x , we conclude that ψ = - e - x - e x + ye - x + h ( y ). Then ψ y = e - x + h 0 ( y ) = N = e - x . Therefore, h 0 ( y ) = 0 and we conclude that the solution is given implicitly by - e - x - e x + ye - x = c . (b)
Image of page 57
74 CHAPTER 2. FIRST ORDER DIFFERENTIAL EQUATIONS 27.(a) Since ( N x - M y ) /M = 1 /y is a function of y only, we know that μ ( y ) = e R 1 /y dy = y is an integrating factor for this equation. Multiplying the equation by μ , we have ydx + ( x - y sin y ) dy = 0 . Then for this equation, M y = 1 = N x . Therefore, this new equation is exact. Integrating M with respect to x , we conclude that ψ = xy + h ( y ). Then ψ y = x + h 0 ( y ) = N = x - y sin y . Therefore, h 0 ( y ) = - y sin y which implies that h ( y ) = - sin y + y cos y , and we conclude that the solution is given implicitly by xy - sin y + y cos y = c . (b) 28.(a) Since ( N x - M y ) /M = (2 y - 1) /y is a function of y only, we know that μ ( y ) = e R 2 - 1 /y dy = e 2 y /y is an integrating factor for this equation. Multiplying the equation by μ , we have e 2 y dx + (2 xe 2 y - 1 /y ) dy = 0 . Then for this equation, M y = N x . Therefore, this new equation is exact. Integrating M with respect to x , we conclude that ψ = xe 2 y + h ( y ). Then ψ y = 2 xe 2 y + h 0 ( y ) = N = 2 xe 2 y - 1 /y . Therefore, h 0 ( y ) = - 1 /y which implies that h ( y ) = - ln y , and we conclude that the solution is given implicitly by xe 2 y - ln y = c or y = e 2 x + ce x + 1.
Image of page 58
Image of page 59

You've reached the end of your free preview.

Want to read all 70 pages?

  • Spring '13
  • YOYL
  • initial condition, First Order Equations, τ dτ

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture