The general solution of $\frac{a \ dx}{(b-c)yz} = \frac{b \ dy}{(c - a)zx} = \frac{c \ dz}{(a - b)xy}$ will be _____, where c1 and c2 are arbitary constants.

Question

The general solution of $\frac{a \ dx}{(b-c)yz} = \frac{b \ dy}{(c - a)zx} = \frac{c \ dz}{(a - b)xy}$ will be _____, where c1 and c2 are arbitary constants.

TuteeHUB · Accepted Answer

ax - by + cz = c1, ax2 + by2 - cz2 = c2

TuteeHUB · Answer

ax2 + by2 - cz2 = c1, a2x2 - b2y2 + c2z2 = c2

TuteeHUB · Answer

ax + by + cz = c1, ax2 = by2 - by2 - cz2 = c2

TuteeHUB · Answer

ax2 + by2 + cz2 = c1, a2x2 + b2y2 + c2z2 = c2

1.	The general solution of \(\frac{a \ dx}{(b-c)yz} = \frac{b \ dy}{(c - a)zx} = \frac{c \ dz}{(a - b)xy}\) will be _____, where c1 and c2 are arbitary constants.
A.	ax2 + by2 - cz2 = c1, a2x2 - b2y2 + c2z2 = c2
B.	ax + by + cz = c1, ax2 = by2 - by2 - cz2 = c2
C.	ax2 + by2 + cz2 = c1, a2x2 + b2y2 + c2z2 = c2
D.	ax - by + cz = c1, ax2 + by2 - cz2 = c2
Answer» D. ax - by + cz = c1, ax2 + by2 - cz2 = c2

The general solution of \(\frac{a \ dx}{(b-c)yz} = \frac{b \ dy}{(c - a)zx} = \frac{c \ dz}{(a - b)xy}\) will be _____, where c1 and c2 are arbitary constants.

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