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Find a family of solutions for the following equation, assume that the coeeficient of dy $\not= 0$ $xy' - y - x sin(\frac{y}{x}) = 0 $ The solution I get when I solve it using the substitution u = y/x is $\sqrt{1- \frac{y^2}{x^2}} + \frac{y}{x} arcsin(\frac{y}{x}) = ln(x) + c$ The answer given inthe book is $ y = 2xarctan(cx)$ I have tried and tried and I can't get the equation to that form. If someone could please please enlighten me.

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This will help you in comparison with you own calculus : $$xy' - y - x \sin(\frac{y}{x}) = 0 $$ $u = y/x$ then $y=xu$ and $y'=xu'+u$ $$x(xu'+u) - xu - x \sin(u) = 0 $$ $$xu' - \sin(u) = 0 $$ $$\frac{du}{\sin(u)}=\frac{dx}{x}$$ $\ln\left(\tan(\frac{u}{2})\right)=\ln(x)$+constant $$\tan(\frac{u}{2})=cx$$ $$u=2\arctan(cx)$$ $$y=2x\arctan(cx)$$

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