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 $x(t)$.
 A: 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 
$$e^{-\int\frac {x}{x^2-x-A}dx}=
C''\left(2abx-a^2-ab\right)^{-\frac{ab+b^2}{2ab}}\left(2abx+a^2-ab\right)^{-\frac{ab-b^2}{2ab}}.$$
A: I have not got an answer but I may have some insight:
Method: Differentiate both equations to get
\begin{align}
\frac{dx}{dt}=a\cos(y)-b\Rightarrow& \frac{d^2x}{dt^2} =-a\sin(y)\frac{dy}{dt} \qquad(1)\newline
x\frac{dy}{dt}=-a\sin(y)\Rightarrow& \frac{dx}{dt}\frac{dy}{dt}+x\frac{d^2y}{dt^2}=-a\cos(y)\frac{dy}{dt}\qquad(2)
\end{align}
I have tried to apply some substitutions between the old and new equations to get some cancellations but can't reach something simple. 
There is only one substitution that seems to lead somewhere.
I use prime notation now to make it readable (where any $f'$ is differentiated with respect to $t$).
We have
$$x'y'+xy''=-a\cos(y)y'\Rightarrow a\cos(y)=\frac{-(x'y'+xy'')}{y'}=-x'-x\frac{y''}{y'}$$
Substituting into the left hand equation of equality $(1)$:
$$x'=-x'-x\frac{y''}{y'}-b\Rightarrow 2x'+x\frac{y''}{y'}=-b $$
Let $v=y'$
$$\Rightarrow2x'+x\frac{v'}{v}=-b\Rightarrow x'+x\frac{v'}{2v}=-\frac{1}{2}b$$
This is a first order linear ODE. We use the integrating factor:
$$\Large\mu=e^{\int\frac{v'}{2v}}=e^{\frac{1}{2}\ln v}=v^{\frac{1}{2}}$$
$$\therefore \left[xv^{\frac{1}{2}}\right]'=-bv^{\frac{1}{2}}\Rightarrow x=v^{-\frac{1}{2}}\int-\frac{1}{2}bv^{\frac{1}{2}}dt$$
$$x(t)=-\frac{1}{2}bv^{-\frac{1}{2}}\int v^{\frac{1}{2}}dt$$
And this is where I end up getting stuck, but even then, I'm not sure if it will lead to $x$ as an explicit function of $t$.
I hope something here leads you to your answer..
