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B
f(yx)=cx
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C
f(yx)=cxy
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D
f(yx)=0
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Solution
The correct option is Bf(yx)=cx We have, xdy=(y+xf(yx)f′(yx))dx ⇒dydx=yx+f(yx)f′(yx) which is homogenous Putting y=vx⇒dydx=v+xdvdx, we obtain v+xdvdx=v+f(v)f′(v)⇒f′(v)f(v)dv=dxx Integrating, we get log f(v)=logx+logc ⇒logf(v)=logcx⇒f(yx)=cx