A) \[-{{x}^{2}}/{{y}^{2}}\]
B) \[-{{y}^{2}}/{{x}^{2}}\]
C) \[{{x}^{2}}/{{y}^{2}}\]
D) \[-{{y}^{2}}/{{x}^{2}}\]
Correct Answer: D
Solution :
[d] Putting \[v=y/x\] so that \[x\frac{dv}{dx}+v=\frac{dv}{dx}\] We have \[x\frac{dv}{dx}+v=v+\phi (1/v)\] \[\Rightarrow \frac{dv}{\phi (1/v)}=\frac{dx}{x};\Rightarrow \log \left| Cx \right|=\int{\frac{dv}{^{\phi (1/v)}}}\] But \[y=\frac{x}{\log \left| Cx \right|}\] is the general solution, So \[\frac{x}{y}=\frac{1}{v}=\log \left| Cx \right|=\int{\frac{dv}{\phi (1/v)}}\] \[\Rightarrow \phi (1/v)=-1/{{v}^{2}}\] (Differentiating w.r.t.v both sides) \[\Rightarrow \phi (x/y)=-{{y}^{2}}/{{x}^{2}}\]
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