Mathematical Methods for Physics and Engineering - Matematica.NET

FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS14.2.2 Exact equationsAn exact first-degree first-order ODE is one of the formA(x, y) dx + B(x, y) dy = 0 and for which ∂A∂y = ∂B∂x . (14.4)In this case A(x, y) dx + B(x, y) dy is an exact differential, dU(x, y) say (seesection 5.3). In other wordsAdx+ Bdy= dU = ∂U ∂Udx +∂x ∂y dy,from which we obtainA(x, y) = ∂U∂x , (14.5)B(x, y) = ∂U∂y . (14.6)Since ∂ 2 U/∂x∂y = ∂ 2 U/∂y∂x we therefore require∂A∂y = ∂B∂x . (14.7)If (14.7) holds then (14.4) can be written dU(x, y) = 0, which has the solutionU(x, y) =c, wherec is a constant and from (14.5) U(x, y) is given by∫U(x, y) = A(x, y) dx + F(y). (14.8)The function F(y) can be found from (14.6) by differentiating (14.8) with respectto y and equating to B(x, y).◮Solvex dy +3x + y =0.dxRearranging into the form (14.4) we have(3x + y) dx + xdy =0,i.e. A(x, y) =3x + y and B(x, y) =x. Since∂A/∂y =1=∂B/∂x, the equation is exact, andby (14.8) the solution is given by∫U(x, y) = (3x + y) dx + F(y) =c 1 ⇒ 3x22 + yx + F(y) =c 1.Differentiating U(x, y) with respect to y and equating it to B(x, y) =x we obtain dF/dy =0,which integrates immediately to give F(y) =c 2 . Therefore, letting c = c 1 − c 2 ,thesolutionto the original ODE is3x 2+ xy = c. ◭2472