A) \[\left( \frac{1}{4},-\frac{1}{2} \right)\]
B) \[\left( -\frac{1}{3},\frac{1}{3} \right)\]
C) \[\left( \frac{1}{4},\frac{1}{2} \right)\]
D) \[\left( \frac{1}{3},-\frac{1}{3} \right)\]
Correct Answer: B
Solution :
\[ydx-xdy-3{{y}^{2}}dy=0\] \[\frac{dx}{dy}=\frac{x}{y}+3y\] \[\frac{dx}{dy}-\frac{x}{y}=3y\] I.f.\[{{e}^{-\int_{{}}^{{}}{\frac{1}{y}dy}}}={{e}^{-\ln y}}=\frac{1}{y}\] \[\therefore \]solution is \[\frac{x}{y}=\int_{{}}^{{}}{3y.\frac{1}{y}dy}\] \[\frac{x}{y}=3y+c\]pass through (1,1) \[\therefore \] \[1=3+c;c=-2\] \[x=3{{y}^{2}}-2y\] (i)\[\left( \frac{1}{4},\frac{1}{2} \right)=\frac{1}{4}=\frac{3}{4}+1\] (ii)\[-\frac{1}{3}=\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}\]
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