A) \[\frac{x\ell nx+y\ell ny}{xy}=C\]
B) \[\frac{x\ell nx-y\ell ny}{xy}=C\]
C) \[\frac{\ell nx}{x}+\frac{\ell ny}{y}\ne C\]
D) \[\frac{\ell nx}{x}-\frac{\ell ny}{y}=C\]
Correct Answer: A
Solution :
| Given, \[\frac{dy}{dx}=\frac{y(x-y\ell ny)}{x(x\ell nx-y)}\]\[\Rightarrow {{x}^{2}}\ell n\,\,x\,\,dy-xy\,\,dy=xy\,\,dx-{{y}^{2}}\ell n\,\,y\,\,dx\] |
| \[\frac{\ell nx}{y}dy-\frac{1}{xy}dy=\frac{1}{xy}dx-\frac{\ell ny}{{{x}^{2}}}dx\](on dividing by \[{{x}^{2}}{{y}^{2}}\]) |
| \[\Rightarrow \frac{1}{xy}dx-\frac{\ell n\,x}{{{y}^{2}}}dy+\frac{1}{xy}dy-\frac{\ell n\,y}{{{x}^{2}}}dx=0\]\[\Rightarrow d\left( \frac{\ell nx}{y} \right)+d\left( \frac{\ell ny}{x} \right)=C\] |
| On integrating both sides, we get \[x\,\ell n\,x+y\,\ell n\,y=Cxy\] |
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