Solve differential equation $y'=a \sin(bt-y)$ I'm looking for a solution to $y'=a \sin(bt-y)$, where a and b are constants.
I've looked at previous posts were b=1, which leads to nice solutions. I'm looking for a solution when b is an arbitrary constant.
To kick off:
Substitute $u=bt-y$
$u'=b-y'$ -> $y'=b-u'$.
$u'=b-a\sin(u)$
$\int\frac{du}{b-a\sin(u)}=\int dt$
And this is where I'm stuck. Differential/integral equation solvers give me some solutions including tans and arctans. Plotting those solutions does not match my numerical approach.
Maybe it has something to do with the domains of the tan and arctan?
I look forward to your help!
 A: Hint.
As
$$
\cases{
\frac{y'}{a}=\sin(bt-y)\\
\frac{y''}{a(b-y')}=\cos(bt-y) 
}\Rightarrow \left(y'\right)^2+\left(\frac{y''}{b-y'}\right)^2=a^2
$$
This ODE has a Wolfram solution.
$$
y = \cot ^{-1}\left(\frac{b \coth \left(\sqrt{a^2-b^2} (t-c_1)\right)}{\sqrt{a^2-b^2}}\right)+b (t-c_1)+c_2\pm 4 \cosh \left(\sqrt{a^2-b^2} (t-c_1)\right) \coth \left(\sqrt{a^2-b^2} (t-c_1)\right) \coth ^{-1}\left(\frac{a
   \sqrt{a^2-b^2}}{a^2-\frac{1}{2} b^2 \text{sech}^2\left(\frac{1}{2} \sqrt{a^2-b^2} (t-c_1)\right)}\right)\sinh ^3\left(\sqrt{a^2-b^2}
   (t-c_1)\right) \text{csch}^2\left(2 \sqrt{a^2-b^2} (t-c_1)\right)i
$$
A: $$y'=a\sin(bt-y)$$
let $u=bt-y\Rightarrow y'=b-u'$ so:
$$b-u'=a\sin(u)$$
$$u'=b-a\sin(u)$$
so:
$$\int dt=\int\frac{du}{b-a\sin(u)}$$
So all agreed up until this point

Now problems of this form often call for the Weierstrauss substitution which would give us:
$$v=\tan(u/2)\\\sin(u)=\frac{2v}{1+v^2}\\du=\frac{2}{1+v^2}dv$$
which would give us:
$$\text{RHS}=\int\frac{1}{b-a\cfrac{2v}{1+v^2}}\frac{2}{1+v^2}dv=\int\frac{dv}{b(1+v^2)-a(2v)}$$
then from here you can simplify the process for yourself by using completing the square
Hope this helps!
