A) 4
B) \[\frac{1}{4}\]
C) -4
D) \[-\frac{1}{4}\]
Correct Answer: D
Solution :
Given\[\frac{dy}{dx}=\frac{y}{x}+\phi \left( \frac{y}{x} \right)\] ?(1) Let\[\left( \frac{y}{x} \right)=v\]so that \[y=xv\]or\[\frac{dy}{dx}=x\frac{dv}{dx}+v\]?(2) from (1) & (2),\[x\frac{dv}{dx}+\cancel{v}=\cancel{v}+\phi \left( \frac{1}{v} \right)\] or,\[\frac{dy}{\phi \left( \frac{1}{v} \right)}=\frac{dx}{x}\] Integrating both sides, we get \[\int_{{}}^{{}}{\frac{dx}{x}=\int_{{}}^{{}}{\frac{dv}{\phi \left( \frac{1}{v} \right)}\Rightarrow \ln x+c=\int_{{}}^{{}}{\frac{dv}{\phi \left( \frac{1}{v} \right)}}}}\](where c being constant of integration) But, given \[y=\frac{x}{\ln |cx|}\]is the general solution so that\[\frac{x}{y}=\frac{1}{v}=\ln |cx|=\int_{{}}^{{}}{\frac{dv}{\phi \left( \frac{1}{v} \right)}}\] Differentiating w.r.t v both sides, we get \[\phi \left( \frac{1}{v} \right)=\frac{-1}{{{v}^{2}}}\Rightarrow \phi \left( \frac{x}{y} \right)=-\frac{{{y}^{2}}}{{{x}^{2}}}\] when\[\frac{x}{y}=2\]i.e.\[\phi (2)=-{{\left( \frac{y}{x} \right)}^{2}}=-{{\left( \frac{1}{2} \right)}^{2}}=\left( \frac{-1}{4} \right)\]
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