Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz2 => Topic started by: Victorwoshinidie on October 19, 2019, 12:26:04 PM

(3x+6/y) + (x^2/y + 3 y/x) dy/dx = 0
We want to find an integrating factor u as a function of xy st. (uM)y = (uN)x, Let z = xy, Thus, u(xy) = u(z(x,y)). Then
ux(xy) = du/dz dz/dx = y du/dz and uy(xy) = du/dz dz/dy = x du/dz
Therefore,
(uM)y = (uN)x => uMy +xMdu/dz = uNx + yNdu/dz
du/dz = u(Nx  My/xM  yN)
Therefore, u(z) = exp(integral R(z) dz ) where R(z) = R(xy) = Nx  My / xM  yN
Returning to our orginal diffrential equation, let
M(x,y) = 3x + 6/y and N(x,y) = x^2/y +3y/x = 0
Then derive both M and N
we get 6/y^2 and 2x/y  3y/x^2
u(xy) = exp( integral 1/z dz) e ^ logz = z = xy
(3x^2y +6x) + (x^3+ 3y ^2) dy/dx = 0
fi (x,y) = x^3y + 3x^2 + y ^3
Thus the solutions of the diffential equation are given implicitly by
x^3y + 3x^2 + y ^3 = C