A) \[(-\sqrt{2},1)\]
B) \[(-1,2)\]
C) \[(\sqrt{3},0)\]
D) \[(3,0)\]
Correct Answer: C
Solution :
Here,\[=\frac{dy}{dx}=\frac{{{x}^{2}}-2y}{x}\]\[\Rightarrow \]\[\frac{dy}{dx}+\frac{2}{x}y=x\] \[\therefore \]\[I.F.={{e}^{\int_{{}}^{{}}{\frac{2}{x}dx}}}={{x}^{2}}\] \[\therefore \]Solutions is\[y.{{x}^{2}}=\int_{{}}^{{}}{x.{{x}^{2}}dx+C}\] \[\Rightarrow \]\[y{{x}^{2}}=\frac{{{x}^{4}}}{4}+C\] Since, the curve passes through the point (\[1,\text{ }-2\]). \[\therefore \]\[-2=\frac{1}{4}+C\Rightarrow C=\frac{-9}{4}\] \[\therefore \]\[y{{x}^{2}}=\frac{{{x}^{4}}}{4}-\frac{9}{4}\]\[\Rightarrow \]\[{{x}^{4}}-4{{x}^{2}}y-9=0\] Now, only the point in option i.e., \[(\sqrt{3},0)\]satisfies the above equation.
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