<s>[INST] Solving system of differential equations to evaluate $x(t)$ I have to solve the following system of equations and find $x(t)$
$$\frac{dx}{dt} =a\cos(y)-b$$
$$x\frac{dy}{dt}=-a\sin(y)$$
Initial conditions:


*

*$t=0,x=x_o, y=\frac{\pi}{2}$


By dividing equations and integration, i obtained $x(y)$:
$$x=x_o \csc(y) \left|\csc(y)-\cot(y)\right|^{\frac{b}{a}}$$
Now I cannot proceed to find $x(t)$.
 [/INST] Too long for a comment. 
Domains of my math competence are far from differential equations, so I can miss some subtleties. I assume that $a$ and $b$ are constants. Given a differentiable function $u=u(t)$ by $u’$ I shall denote its derivative with respect to $t$. 
Differentiating the first equation we obtain 
$x’’=-a\sin y\cdot y’.$
Then 
$xx’’=-a\sin y\cdot xy'=a^2\sin^2 y=a^2-(x’+b)^2=a^2-(x’)^2-2bx’-b^2$
$(xx’)’=xx’’+(x’)^2=a^2 -b^2-2bx’$
Integrating with respect to $t$ we obtain  
$xx’=(a^2 -b^2)t-2bx+C$, 
This equation looks simpler than the initial system and maybe it already can be solved by standard methods. 
Also we can find $C$ as follows. Applying the initial conditions we obtain 
$x(0)x’(0)=(a^2 -b^2)\cdot 0 -2bx(0)+C$, 
but $x(0)=x_o$ and $x’(0)=a\cos y(0)-b= a\cos\frac\pi2-b=-b$, so $-bx_o=-2bx_o+C$, 
and $C=bx_o$. 
Update. According to player100 comment, we have to calculate $\int\frac {x}{x^2-x-A}dx$. But $$x^2-x-A= x^2-x+\frac{b^2-a^2}{4b^2}.$$ The discriminant of this polynomial is $$1-4\frac{b^2-a^2}{4b^2}=\frac{a^2}{b^2}.$$ Thus 
$$x^2-x-A=\left(x-\frac{b+a}{2b}\right)\left(x-\frac{b-a}{2b}\right).$$
If $a=0$ then 
$$\frac x{x^2-x-A}=\frac{x}{(x-\frac 12)^2}=\frac{1}{x-\frac 12}+\frac{1}{2(x-\frac 12)^2}. $$
So
$$\int\frac {x}{x^2-x-A}dx=\int \frac{dx}{x-\frac 12}dx+\int\frac{dx}{2(x-\frac 12)^2}dx=\ln\left(x-\frac 12\right)-\frac 1{2x-1}+C’$$
(provided $x>\frac 12$).
Then 
$$e^{-\int\frac {x}{x^2-x-A}dx}=C’’\frac 1{x-\frac 12}e^{\frac 1{2x-1}}.$$
If $a\ne 0$ then 
$$\frac x{x^2-x-A}=\frac{ab+b^2}{2abx-a^2-ab}+
\frac{ab-b^2}{2abx+a^2-ab}.$$
So 
$$\int\frac {x}{x^2-x-A}dx=$$ $$\int \frac{ab+b^2}{2abx-a^2-ab}dx+
\int\frac{ab-b^2}{2abx+a^2-ab}dx=$$ $$\frac{ab+b^2}{2ab}\ln(2abx-a^2-ab)+\frac{ab-b^2}{2ab}\ln(2abx+a^2-ab)+C'.$$
(provided the denominators are positive)
Then 
$$e^{-\int\frac {x}{x^2-x-A}dx}=
C''\left(2abx-a^2-ab\right)^{-\frac{ab+b^2}{2ab}}\left(2abx+a^2-ab\right)^{-\frac{ab-b^2}{2ab}}.$$
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