A) \[(x-y){{(x+2y)}^{2}}=c\]
B) \[y=x+c\]
C) \[y=(2y-x)+c\]
D) \[y=\frac{x}{2y-x}+c\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}=\frac{x}{2y-x}\]. Put \[y=vx\]Þ\[v+x\frac{dv}{dx}=\frac{dy}{dx}\] \[v+x\frac{dv}{dx}=\frac{x}{2v-x}=\frac{1}{2v-1}\] \[x\frac{dv}{dx}=\frac{1}{2v-1}-v=\frac{1-2{{v}^{2}}+v}{2v-1}=-\frac{(v-1)(2v+1)}{2v-1}\] \[\frac{(2v-1)}{(2v+1)(v-1)}=\frac{-dx}{x}\]; \[\frac{1}{3(v-1)}+\frac{4}{3(2v+1)}=\frac{-dx}{x}\] \[\frac{1}{3}\log (v-1)+\frac{4}{3}.\frac{1}{2}\log (2v+1)=\log \frac{1}{x}+\log c\] \[\log {{(v-1)}^{1/3}}+\log {{(2v+1)}^{2/3}}=\log \frac{c}{x}\] \[={{(v-1)}^{1/3}}{{(2v+1)}^{2/3}}=\frac{c}{x}\] \[\left( \frac{y-x}{x} \right)\text{ }{{\left( \frac{2y+x}{x} \right)}^{2}}=\frac{{{c}^{3}}}{{{x}^{3}}}\] Þ \[(x-y){{(x+2y)}^{2}}=c\].
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