A) \[{{x}^{2}}+\frac{{{y}^{2}}}{2}+\frac{1}{3}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{3/2}}=C\]
B) \[x-\frac{{{y}^{2}}}{3}+\frac{1}{2}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{1/2}}=C\]
C) \[x-\frac{{{y}^{2}}}{2}+\frac{1}{3}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{3/2}}=C\]
D) None of these
Correct Answer: C
Solution :
[c] Rearranging the equation, we have \[dx-ydy+\sqrt{\left( {{x}^{2}}+{{y}^{2}} \right)}(xdx+ydy)=0\] \[\Rightarrow dx-ydy+\frac{1}{2}\sqrt{\left( {{x}^{2}}+{{y}^{2}} \right)}d({{x}^{2}}+{{y}^{2}})=0\] On integrating, we get \[x-\frac{{{y}^{2}}}{2}+\frac{1}{2}\int{\sqrt{t}dt=c,\left\{ t=\sqrt{\left( {{x}^{2}}+{{y}^{2}} \right)} \right\}}\] Or \[x-\frac{{{y}^{2}}}{2}+\frac{1}{3}{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{3/2}}=c\]
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