A) \[xy\cos \left( \frac{y}{x} \right)=K\]
B) \[\cos \left( \frac{x}{y} \right)=K\,xy\]
C) \[\frac{x}{y}\cos \left( \frac{y}{x} \right)=K\]
D) None of these
Correct Answer: A
Solution :
The given differential equation is \[y\left[ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right]dx\] \[-x\left[ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right]dy=0\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{y\left[ x\cos \left( \frac{y}{x} \right)+y\sin \left( \frac{y}{x} \right) \right]}{x\left[ y\sin \left( \frac{y}{x} \right)-x\cos \left( \frac{y}{x} \right) \right]}\] Putting \[y=vx,\] \[\Rightarrow \] \[\frac{dy}{dx}=v+x\frac{dv}{dx}\]in Eq. (i), we get \[v+x\frac{dv}{dx}=\frac{vx[x\cos v+vx\sin v]}{x[vx\sin v-x\cos v]}\] \[\Rightarrow \] \[x\frac{dv}{dx}=\frac{vx\cos v+{{v}^{2}}x\sin v}{vx\sin v-x\cos v}-v\] \[\Rightarrow \]\[x\frac{dv}{dx}\] \[=\frac{vx\operatorname{cosv}+{{v}^{2}}x\sin v-{{v}^{2}}x\sin v+xv\cos v}{vx\sin v-x\cos v}\] \[\Rightarrow \]\[x\frac{dv}{dx}=\frac{2xv\cos v}{xv\sin v-x\cos v}\] \[\Rightarrow \]\[\frac{v(\sin v-\cos v)dv}{v\cos v}=2\frac{dx}{x}\] Integrating both sides, we get \[-\int_{{}}^{{}}{\frac{(\cos v-v\sin v)dv}{v\cos v}}=2\int_{{}}^{{}}{\frac{dx}{x}}\] \[\Rightarrow \]\[-\log |v\cos v|=2\log |x|+\log c\] \[\Rightarrow \]\[\log \left| \frac{1}{v\cos v} \right|=\log |{{x}^{2}}|+\log c\] \[\Rightarrow \]\[\frac{1}{v\cos v}=c{{x}^{2}}\] \[\Rightarrow \]\[\frac{x}{y}\sec \left( \frac{y}{x} \right)=c{{x}^{2}}\] \[\Rightarrow \]\[xy\cos \left( \frac{y}{x} \right)=\frac{1}{c}\] \[\Rightarrow \]\[xy\cos \left( \frac{y}{x} \right)=k,\left( k=\frac{1}{c} \right)\]
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