ordinary differential equations - How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? -
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Solved] Solve : yz^2 (x^2 - yz ) dx +zx^2 (y^2 -xz ) dy +xy^2 (z^2 -xy ) dz... | Course Hero
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SOLVED: Rewrite the iterated integral 1-x 2-2y J' J' Jo f(c,y,2) dz dy dx changing the order of integration to first with respect to X, then Y, and then z. y-1 'K "
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Solved] Solve : yz^2 (x^2 - yz ) dx +zx^2 (y^2 -xz ) dy +xy^2 (z^2 -xy ) dz... | Course Hero
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