A) \[x\phi \left( \frac{y}{x} \right)=k\]
B) \[\phi \left( \frac{y}{x} \right)=kx\]
C) \[y\phi \left( \frac{y}{x} \right)=k\]
D) \[\phi \left( \frac{y}{x} \right)=ky\]
E) \[\phi \left( \frac{y}{x} \right)=k{{e}^{y/x}}\]
Correct Answer: B
Solution :
The given differential equation can be written as \[\frac{dy}{dx}-\frac{y}{x}=\frac{\phi \left( \frac{y}{x} \right)}{\phi \left( \frac{y}{x} \right)}\] \[\Rightarrow \] \[x\phi \left( \frac{y}{x} \right)\left( \frac{1}{x}\frac{dy}{dx}-\frac{y}{{{x}^{2}}} \right)=\phi \left( \frac{y}{x} \right)\] \[\Rightarrow \] \[\frac{\phi \left( \frac{y}{x} \right)\left( \frac{x\frac{dy}{dx}-y}{{{x}^{2}}} \right)}{\phi \left( \frac{y}{x} \right)}=\frac{1}{x}\] \[\Rightarrow \] \[\int{\frac{\phi \left( \frac{y}{x} \right)d\left( \frac{y}{x} \right)}{\phi \left( \frac{y}{x} \right)}}=\int{\frac{1}{x}}dx+\log k\] \[\Rightarrow \] \[\log \phi \left( \frac{y}{x} \right)=\log x+\log k\] \[\Rightarrow \] \[\phi \left( \frac{y}{x} \right)=kx\]
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