Substituting y = vx in Differential EquationsDate: 02/16/2001 at 06:20:44 From: Andrew Thomson Subject: Differential equations This equation is supposed to be solvable by separating the variables, but I can't figure it out. (y^2 - x^2)dy + 2xy*dx = 0 Date: 02/16/2001 at 15:19:08 From: Doctor Anthony Subject: Re: Differential equations Try putting your equation in this form: dy/dx = 2xy/(x^2 - y^2) The right-hand side can be written as a function of y/x: dy/dx = 2(y/x)/(1 - (y/x)^2) = f(y/x) For an equation of the form dy/dx = f(y/x), making the substitution y = vx is guaranteed to render it separable. (An easier test for this condition is to see whether f(x,y) = f(vx,vy).) The product rule gives dy/dx = v + x(dv/dx) Plug this into your equation: (v^2*x^2 - x^2)(v + x(dv/dx)) + 2x^2*v = 0 Divide through by x^2: (v^2 - 1)(v + x(dv/dx)) + 2v = 0 v + x(dv/dx) = -2v/(v^2-1) x(dv/dx) = -2v/(v^2-1) - v x(dv/dx) = [-2v - v(v^2-1)]/(v^2-1) x(dv/dx) = -v(2 + v^2 - 1)/(v^2-1) x(dv/dx) = -v(v^2+1)/(v^2-1) (v^2-1)dv dx --------- = - -- v(v^2+1) x -1 2v dx [-- + -----]dv = - -- v v^2+1 x Integrate: -ln(v) + ln(v^2+1) = - ln(x) + C ln[(v^2+1)/v] = ln(1/x) + ln(A) where A is a constant (v^2+1)/v = A/x v + 1/v = A/x Substitute v = y/x: y/x + x/y = A/x y^2 + x^2 A --------- = - xy x y^2 + x^2 = Ay x^2 + y^2 - Ay = 0 which is the equation of a circle. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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