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I need help with solving this differential equation taken from Ordinary Differential Equations by Tenenbaum and Pollard.

$$ xy'-y-x\sin(y/x)=0 $$

I used the subsitution, $u=y/x$ to obtain,

$$x^2\,du-x\sin(u)\,dx=0$$ $$\csc(u)\,du-\frac{1}{x}\,dx=0$$ Then by integrating, $$\ln|\csc(u)+\cot(u)|+\ln|x|=C$$ $$|\csc u+\cot u||x|=D$$ and rewriting $u$, $$\left|\frac{1+\cos (y/x)}{\sin (y/x)}\right||x|=D\quad (x\neq0\,,\frac{y}{x}\neq \pm n\pi)$$

However, the provided solution is $y=2x\arctan cx$. How do I transform the solution I obtained, and how do I deal with the absolute values?

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The next step is

$$\frac{1+\cos(u)}{\sin(u)}=\frac{2\cos^2(\frac u2)}{2\sin(\frac u2)\cos(\frac u2)}=\cot(\frac u2).$$

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    $\begingroup$ Rather cotan(u/2). $\endgroup$ – Did Jul 17 '15 at 9:36
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    $\begingroup$ Yeah, and 1+cos(u) is... $\endgroup$ – Did Jul 17 '15 at 9:43
  • $\begingroup$ @did: ok like this. $\endgroup$ – Yves Daoust Jul 17 '15 at 9:58
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You properly arrived to $$x \frac{du}{dx}=\sin(u)$$ which is effectively separable and gives $$\frac{dx}x=\frac{du}{\sin(u)}$$ So $$\int\frac{dx}x=\int\frac{du}{\sin(u)}$$ The lhs does not make any problem; for the rhs, use the tangent half-angle substitution $t=\tan(\frac u 2)$ to get $$\int\frac{du}{\sin(u)}=\int\frac{dt}{t}$$ So, $$\log(x)+a=\log(t)=\log(cx)$$ $$t=cx=\tan(\frac u 2)$$ $$u=2 \tan^{-1}(cx)$$ $$y=ux=2\, x\,\tan^{-1}(cx)$$

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