Author Topic: Quiz2 TUT0702  (Read 470 times)

Victorwoshinidie

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Quiz2 TUT0702
« 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