The solution of xdy=y+xf(y/x)f1(y/x)dx is

xdy=y+xf(y/x)f1(y/x)dx ⇒xdy-ydxx2=f(y/x)f1(y/x)dxx ⇒dyx=f(y/x)f1(y/x)dxx ⇒f1(y/x)f (y/x)dyx=dxx Integrate ,then we will get log f(y/x)=log xc

The solution of xdy=y+xf(y/x)f1(y/x)dx is

xdy=y+xf(y/x)f1(y/x)dx ⇒xdy-ydxx2=f(y/x)f1(y/x)dxx ⇒dyx=f(y/x)f1(y/x)dxx ⇒f1(y/x)f (y/x)dyx=dxx Integrate ,then we will get log f(y/x)=log xc

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The solution of $xdy = [y + x \frac{f (y / x)}{f^{1} (y / x)}] dx$ is

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$| f (y / x) | = c | x |, c > 0$

$| f (y / x) | = |\frac{y}{x}| + c, c > 0$

$| f (y / x) | = |\frac{y}{x} + x| + c$

$| f (y / x) | = |\frac{x}{y} + c|, c > 0$

answer is A.

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Detailed Solution

$xdy = [y + x \frac{f (y / x)}{f^{1} (y / x)}] dx \Rightarrow \frac{xdy - y d x}{x^{2}} = \frac{f (y / x)}{f^{1} (y / x)} \frac{dx}{x} \Rightarrow d (\frac{y}{x}) = \frac{f (y / x)}{f^{1} (y / x)} \frac{dx}{x} \Rightarrow \frac{f^{1} (y / x)}{f^{} (y / x)} d (\frac{y}{x}) = \frac{dx}{x} I n t e g r a t e, t h e n w e w i l l g e t \log |f (y / x)| = \log |x c|$