Differential Equations Question 189
Question: Solution of the differential equation $ (y+x\sqrt{xy}(x+y))dx+(y\sqrt{xy}(x+y)-x)dy=0 $ is
Options:
A) $ \frac{x^{2}+y^{2}}{2}+{{\tan }^{-1}}\sqrt{\frac{y}{x}=c} $
B) $ \frac{x^{2}+y^{2}}{2}+2{{\tan }^{-1}}\sqrt{\frac{x}{y}=c} $
C) $ \frac{x^{2}+y^{2}}{2}+2{{\cot }^{-1}}\sqrt{\frac{x}{y}=c} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The given equation is written as $ ydx-xdy+x\sqrt{xy}(x+y)dx+y\sqrt{xy}(x+y)dy=0 $ or $ ydx-xdy+(x+y)\sqrt{xy}(xdx+ydy)=0 $
Or $ \frac{ydx-xdy}{y^{2}}+( \frac{x}{y}+1 )\sqrt{\frac{x}{y}}( d( \frac{x^{2}+y^{2}}{2} ) )=0 $
or $ d( \frac{x^{2}+y^{2}}{2} )+\frac{d( \frac{x}{y} )}{( \frac{x}{y}+1 )\sqrt{\frac{x}{y}}}=0 $
Or $ \frac{x^{2}+y^{2}}{2}+2{{\tan }^{-1}}\sqrt{\frac{x}{y}}=c $
BETA
