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Question: The solution of the equation \(\left( y + x\sqrt{xy}(x + y) \right)dx - \left( y + y\sqrt{xy}(x + y...

The solution of the equation

(y+xxy(x+y))dx−(y+yxy(x+y))dy=0\left( y + x\sqrt{xy}(x + y) \right)dx - \left( y + y\sqrt{xy}(x + y) \right)dy = 0is

A

x2+y2=2tan⁡−1yx+cx^{2} + y^{2} = 2\tan^{- 1}\sqrt{\frac{y}{x}} + c

B

x2+y2=4tan⁡−1yx+cx^{2} + y^{2} = 4\tan^{- 1}\sqrt{\frac{y}{x}} + c

C

x2+y2=tan⁡−1yx+cx^{2} + y^{2} = \tan^{- 1}\sqrt{\frac{y}{x}} + c

D

None of these

Answer

x2+y2=4tan⁡−1yx+cx^{2} + y^{2} = 4\tan^{- 1}\sqrt{\frac{y}{x}} + c

Explanation

Solution

The given equation can be written as

x dx + y dy + ydx−xdyxy(x+y)=0\frac{ydx - xdy}{\sqrt{xy}(x + y)} = 0

or 12d(x2+y2)=xdy−ydxx2yx(1+yx)=2(1+yx)d(yx)\frac{1}{2}d(x^{2} + y^{2}) = \frac{xdy - ydx}{x^{2}\sqrt{\frac{y}{x}}\left( 1 + \frac{y}{x} \right)} = \frac{2}{\left( 1 + \frac{y}{x} \right)}d\left( \sqrt{\frac{y}{x}} \right)

.⇒ x2+y2=4tan⁡−1yx+cx^{2} + y^{2} = 4\tan^{- 1}\sqrt{\frac{y}{x}} + c