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So I have to solve the homog. diff. equation $$xy'=\left[x \sin\left(\frac{y}{x}\right)\right]+y$$. now,I know how to prove that it is a homogeneous diff equation of degree 0 ,so I won't write this one.

To the solution : $u=\frac yx$

$$y'=\frac{x\sin\left(\frac yx\right)+y}x$$ here $$u'x +u=\frac{x\sin\left(\frac{ux}x\right)+ux}x=\sin u+u$$ $$u'x=\sin u$$ $$\int \frac 1{\sin u}du=\int \frac 1x dx$$ $$\ln|\sin u|=\ln|xc|$$ $$\sin u=xc$$

When I replace $\;u=\frac yx$ I dont get $\;y=2\;x\arctan(c\,x)$ which is the answer in my book..why?

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Could you clean up the presentation of your question? –  alex.jordan May 6 '13 at 21:36
    
what do you mean?? –  gnsf May 6 '13 at 21:36
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Use LaTeX better. I do not want to spend the effort to try to read the ASCII version of the question. –  alex.jordan May 6 '13 at 21:37
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For a $\LaTeX$ primer, you can see this. If you must use text, please add some line breaks and be careful with parentheses. Note the last $\ln |x*c|\sin u=x*c$ looks wrong. –  Ross Millikan May 6 '13 at 21:41

1 Answer 1

Everything was right up to : $$\int \frac{du}{\sin(u)}=C+\int\frac{dx}x$$ Your error is in the integration at the left that should return : $$\ \log\left|\tan\left(\frac u2\right)\right|=C+\log|x|$$ Can you finish now?

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Yes.thank you mate ! –  gnsf May 6 '13 at 21:39
    
You are welcome @gnsf (btw I edited your question). –  Raymond Manzoni May 6 '13 at 21:52

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