A) \[\log (y-x)=c+\frac{y-x}{x}\]
B) \[\log (y-x)=c+\frac{x}{y-x}\]
C) \[y-x=c+\log \frac{x}{y-x}\]
D) \[y-x=c+\frac{x}{y-x}\]
Correct Answer: B
Solution :
Given \[x+y\frac{dy}{dx}=2y\]Þ\[\frac{x}{y}+\frac{dy}{dx}=2\] Put \[y=vx\]and \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\therefore \frac{1}{v}+v+x\frac{dv}{dx}=2\] Þ \[v+x.\frac{dv}{dx}=\frac{2v-1}{v}\] Þ \[\frac{v}{{{(v-1)}^{2}}}dv=-\frac{dx}{x}\] Þ \[\frac{v-1+1}{{{(v-1)}^{2}}}dv=-\frac{dx}{x}\] \[\left[ \frac{1}{(v-1)}+\frac{1}{{{(v-1)}^{2}}} \right]dv=-\frac{dx}{x}\] Integrating both sides, \[\int_{{}}^{{}}{\frac{dv}{v-1}}+\int_{{}}^{{}}{\frac{dv}{{{(v-1)}^{2}}}}=-\int_{{}}^{{}}{\frac{dx}{x}}\] Þ \[\log (v-1)-\frac{1}{v-1}=-\log x+c\]Þ \[\log (y-x)=\frac{x}{y-x}+c\].
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