A) \[\phi \left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=cx\]
B) \[x\phi \left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=c\]
C) \[\phi \left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=c{{x}^{2}}\]
D) \[{{x}^{2}}\phi \left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=c\]
Correct Answer: C
Solution :
[c]: We have, \[y\frac{dy}{dx}=x\left[ \frac{{{y}^{2}}}{{{x}^{2}}}+\frac{\phi \left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)}{\phi '\left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)} \right]\] Putting \[\frac{y}{x}=v\Rightarrow y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx},\]we get \[vx\left( v+x\frac{dv}{dx} \right)=x\left[ {{v}^{2}}+\frac{\phi ({{v}^{2}})}{\phi '({{v}^{2}})} \right]\] \[\Rightarrow \]\[{{v}^{2}}+vx\frac{dv}{dx}={{v}^{2}}+\frac{\phi ({{v}^{2}})}{\phi '({{v}^{2}})}\] \[\Rightarrow \]\[v\frac{\phi '({{v}^{2}})}{\phi ({{v}^{2}})}dv=\frac{dx}{x}\] On integrating, we get\[\int_{{}}^{{}}{\frac{dx}{x}}=\int_{{}}^{{}}{\frac{v\phi '({{v}^{2}})}{\phi ({{v}^{2}})}}dv\] \[\Rightarrow \]\[\ln |x|+\ln |{{c}_{1}}|=\frac{1}{2}\ln |\phi ({{v}^{2}})|\] \[\Rightarrow \]\[\ln |{{c}_{1}}x|=\ln |\phi ({{v}^{2}}){{|}^{1/2}}\] \[\Rightarrow \]\[\ln |{{c}_{1}}x|=\phi ({{v}^{2}}){{|}^{1/2}}\Rightarrow c_{1}^{2}{{x}^{2}}=\phi ({{v}^{2}})\] \[\Rightarrow \]\[c{{x}^{2}}=\phi \left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)\][where\[c_{1}^{2}=c\]]
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