A) \[y(1-xy)=Ax\]
B) \[{{y}^{3}}-x=Ay\]
C) \[x(1-xy)=Ay\]
D) \[x(1+xy)=Ay\] Where A is any arbitrary constant
Correct Answer: B
Solution :
\[(x+2{{y}^{3}})\frac{dy}{dx}=y\] Þ \[\frac{dy}{dx}=\frac{y}{x+2{{y}^{3}}}\] Þ \[\frac{dx}{dy}=\frac{x+2{{y}^{3}}}{y}\] or \[\frac{dx}{dy}-\frac{x}{y}=2{{y}^{2}}\], which is a linear equation of the form \[\frac{dx}{dy}+Px=Q\] So, integrating factor (I.F.)\[={{e}^{-\int_{{}}^{{}}{\frac{1}{y}dy}}}\]and solution is \[x\frac{1}{y}=\int_{{}}^{{}}{\frac{1}{y}2{{y}^{2}}dy+A={{y}^{2}}+A}\] Þ \[x={{y}^{3}}+Ay\] Þ \[{{y}^{3}}-x=Ay;\]where A can be \[-ve\]or \[+ve\].
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