MAT244--2019F > Quiz-2

Quiz2 TUT0702

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**Victorwoshinidie**:

(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

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