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I just got started on differential equations and the initial value problem and solved some simpler IVPs in the form of for example $y{\prime} = \frac{2x}{(1+x^2)y}, y(0) = -2$. Now I'm stuck at the problem: $y^{\prime} = cos(y)x, y(0) = 0$, as I'm able to integrate the seperable equation and get $ln (|sec(y) + tan(y)|) = \frac{x^2}{2}+C$ with $C$ being a constant, but I do not know how to solve for $y$ and fullfill the initial value of $y(0) = 0$. If anybody could help me with this, I'd be very thankful.

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    $\begingroup$ $y$ has disappeared altogether in your answer. Did you mean $\log|\sec y+\tan y|$? In any event, you get $C$ by putting in $x=0$ and $y=0$. $\endgroup$ – Gerry Myerson Jan 13 '20 at 2:20
  • $\begingroup$ Of course, I just edited it. Thanks for pointing it out. $\endgroup$ – psyph Jan 13 '20 at 2:21
  • $\begingroup$ $\ln(|\sec(0)+\tan(0)|)=0^2+C $ You get $C=0$ since $\ln 1 =0$ $\endgroup$ – Aryadeva Jan 13 '20 at 2:24
  • $\begingroup$ @Isham I dont think I can do that at this point since I don't know that y(x) looks like, yet... $\endgroup$ – psyph Jan 13 '20 at 3:23
  • $\begingroup$ Oh yes you can do it. Do you know what an implicit form is ? You don't need an explicit form for y to find the constant c. Sometimes it's hard to find such an explicit form and we leave the answer in implicit form psyph $\endgroup$ – Aryadeva Jan 13 '20 at 3:25
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This identity might be useful:

$\tan\left(\frac{\theta}{2}+\frac{\pi}{4} \right)=\sec \theta + \tan \theta$

For more details see: https://en.wikipedia.org/wiki/List_of_trigonometric_identities

Also, shouldn't the left side be functions of $y$ in your solution?

I hope this helps.

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  • $\begingroup$ Thanks, you are right, I made a typo. And yes, that identity helps me a lot. $\endgroup$ – psyph Jan 13 '20 at 2:23
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What is amazing in this problem is that we can use the tangent half-angle substition.

Let $$y=2 \tan ^{-1}(z)\implies y'=\frac{2 z'}{z^2+1}$$ which make the equation to be $$2 z'+x \left(z^2-1\right)=0\implies \log(z^2-1)=-x+c\implies 2 \tanh ^{-1}(z)=-x+c$$ whih finally makes $$y=2 \tan ^{-1}\left(\tanh \left(c_1+\frac{x^2}{4}\right)\right)$$ and the condition leads to $c_1=0$.

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