Question

Find the solution of the general equation of the differential equation:

(1-cosx)y' - ysinx =0, x ≠ k2π

Answer 1

Notice that the condition x ≠ 2πk for (presumably) integer k means cos(x) ≠ ±1, and in particular cos(x) ≠ 1 so that we could divide both sides by (1 - cos(x)) safely. Doing so lets us separate the variables:

(1 - cos(x)) y' - y sin(x) = 0

==> (1 - cos(x)) y' = y sin(x)

==> y'/y = sin(x)/(1 - cos(x))

==> dy/y = sin(x)/(1 - cos(x)) dx

Integrate both sides and solve for y. On the right, substitute u = 1 - cos(x) and du = sin(x) dx.

∫ dy/y = ∫ sin(x)/(1 - cos(x)) dx

∫ dy/y = ∫ du/u

ln|y| = ln|u| + C

exp(ln|y|) = exp(ln|u| + C )

exp(ln|y|) = exp(ln|u|) exp(C )

y = Cu

y = C (1 - cos(x))