A) \[{{x}^{2}}(2xy+{{y}^{2}})={{c}^{2}}\]
B) \[{{x}^{2}}(2xy-{{y}^{2}})={{c}^{2}}\]
C) \[{{x}^{2}}({{y}^{2}}-2xy)={{c}^{2}}\]
D) None of these
Correct Answer: A
Solution :
It can be written in the form of homogeneous equation \[\frac{dy}{dx}=-\frac{3xy+{{y}^{2}}}{{{x}^{2}}+xy}\] So, now put \[y=vx\]and \[\frac{dy}{dx}=v+x\frac{dv}{dx},\]we get \[v+x\frac{dv}{dx}=-\frac{3{{x}^{2}}v+{{x}^{2}}{{v}^{2}}}{{{x}^{2}}+{{x}^{2}}v}\]Þ\[x\frac{dv}{dx}=\frac{-2v(v+2)}{v+1}\] Þ \[\frac{1}{x}dx=-\frac{v+1}{2v(v+2)}dv=-\left[ \frac{1}{2(v+2)}+\frac{1}{2v(v+2)} \right]\,dv\] Þ \[-\frac{2}{x}dx=\left[ \frac{1}{v+2}+\frac{1}{2v}-\frac{1}{2v(v+2)} \right]\] On integrating, we get \[-2{{\log }_{e}}x=\frac{1}{2}\log (v+2)+\frac{1}{2}\log v+\log c\] Þ \[v(v+2){{x}^{4}}={{c}^{2}}\] Þ \[\frac{y}{x}\left( \frac{y}{x}+2 \right){{x}^{4}}={{c}^{2}}\], \[\left( \because v=\frac{y}{x} \right)\] Hence required solution is \[({{y}^{2}}+2xy){{x}^{2}}={{c}^{2}}\].
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