Initial value problem of the form $y^{\prime} = cos(y)x, y(0) = 0$

Question

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.

Answer 1

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.

Answer 2

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 $c_1=0$.