A) \[{{y}^{2}}=-2x\]
B) \[y=-2x\]
C) \[{{y}^{3}}=-2x\]
D) None of these
Correct Answer: C
Solution :
[c] \[(x{{y}^{3}}-{{x}^{2}})dy-(xy+{{y}^{4}})dx=0\] \[\Rightarrow {{y}^{3}}(xdy-ydx)-x(xdy+ydx)=0\] \[\Rightarrow {{x}^{2}}{{y}^{3}}\frac{(xdy-ydx)}{{{x}^{2}}}-x(xdy+ydx)=0\] \[\Rightarrow {{x}^{2}}{{y}^{3}}d\left( \frac{y}{x} \right)-xd(xy)=0\] Dividing by \[{{x}^{3}}{{y}^{2}},\] we get\[\frac{y}{x}d\left( \frac{y}{x} \right)-\frac{d(xy)}{{{x}^{2}}{{y}^{2}}}=0\] Now, integrating \[\frac{1}{2}{{\left( \frac{y}{x} \right)}^{2}}+\frac{1}{xy}=c\] It passes through the point \[(4,-2).\] \[\Rightarrow \frac{1}{8}-\frac{1}{8}=c\Rightarrow c=0\therefore {{y}^{3}}=-2x\]
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