Solve the differential equation csc(x)*dy/dx=exp(-y), given that y(0)=0. (Typical Math HL paper 3 question, Calculus optional topic)

Firstly, we need to grasp the problem, and understand what the question is asking us to do. We are given a differential equation, and we are required to solve for y, i.e. write y as a function of x: y(x)=... . The equation has initial conditions that we must impose, which means we need to make sure that y is zero when x is zero y(x=0)=0.
Now the solution. Recognising that this is a separable ODE, which means the equation can be written as some function of y equals some function of x: f(x)=g(y). Note that csc(x)=1/sin(x), a trigonometric identity which can be found in the formula booklet. In order to move all terms of y to one side and terms of x to the other side, we can multiply through by sin(x) and dx (treat differential as a fraction), and divide by exp(-y), giving us dy/exp(-y)=sin(x)dx. This should ring a bell; since we have differentials, we should now integrate them, recognizing that 1/exp(-y)=exp(y) by rules of exponents. ∫exp(y)dy= ∫sin(x)dx. These are standard integrals, and using the formula booklet, we obtain the result: exp(y)=-cos(x)+c. Do not forget c, as it is the integration constant. We want to write y as a function of x, so we need to solve for y, which we can do by applying the inverse natural logarithm operation on both sides, which is inverse of exp: y(x)=ln[c-cos(x)]. The equation is almost solved, now we need to impose initial conditions. To do that we simply substitute for y and x as asked by the question, in this case y=x=0, and solve for c: 0=ln[c-1], c-1=exp(0)=1. =>c=2. Hence, our final solution is y(x)=ln[2-cos(x)].