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How can I solve $y-xy'-\sin(y')=0$? Are there any general techniques for solving ODE of the form $y=f(y')$, where $f$ is a trigonometric function?

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3 Answers 3

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Assuming you mean $y(x)$, what you have is what is known as a Clairaut's equation.

Separate your $y$ and $y'$ and then differentiate with respect to $x$, then factor and solve.

After a bit of manipulation, you should get $y = \sin \alpha + \alpha x$ or $y = \pm(x \arccos (-x) + \sqrt{-x^2 +1})$.

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  • $\begingroup$ Thanks to you, I remember NOW what I learnt 55 years ago. $\endgroup$ Nov 23, 2013 at 9:22
  • $\begingroup$ I think here it must be $arcsin$ $\endgroup$
    – hhsaffar
    Nov 23, 2013 at 9:39
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I really don't know how to solve this kind of ODE's in a general manner. For your specific case, basically by trial and error, I found y = a x + Sin[a]. Please, let me know if you receive answers. I am really curious.

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May be this helps

$y−xy′−sin(y′)=0 \to sin(y')=y-xy'\\ y'-(xy''+y')-y''cos(y')=0 \to cos(y')=-\frac{xy''}{y''}=-x\\ sin^2(y')+cos^2(y')=1 \to (y-xy')^2+x^2=1 \to y^2+x^2(y')^2-2xyy'+x^2=1$

Which is a first-order nonlinear ordinary differential equation, that must be solvable.

Let $y=xv(x)$.

After doing the math we will get: $y(x) = x (-sin^{-1}(x)-\frac{\sqrt{-x^2+1}}{x}+c_1)$ or $y(x) = x (sin^{-1}(x)+\frac{\sqrt{-x^2+1}}{x}+c_1)$

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  • $\begingroup$ Please check the answer again, I didn't do final check. $\endgroup$
    – hhsaffar
    Nov 23, 2013 at 9:33

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