how could i solve this differential equation?

Question

square (h) +[hsin(theta)* dh/d(theta)] =1
this can be read as : h square + h into sin theta * differential of h w.r.t. theta = 1
here h is function of theta
h= h(theta)

Answer 1

Let u=h^2.

Then du/d(theta) = 2h*dh/d(theta), and the differential equation becomes

u + (1/2) sin(theta) du/d(theta) = 1.

This is a first-order linear ODE, which can be solved using an integrating factor. It's going to get a bit messy, since you'd need to write it as

du/d(theta) + 2 csc(theta) u = 2 csc(theta)

and use the integrating factor

g(theta) = exp(2*antiderivative of csc theta)
= exp(-2 ln |csc(theta) + cot(theta)|)
= 1/|csc(theta) + cot(theta)|^2.

You end up with

d(u/|csc(theta)+cot(theta)|^2) / d(theta) =
2 csc(theta) / |csc(theta)+cot(theta)|^2

so

u(theta) = |csc(theta)+cot(theta)|^2 *
antiderivative of csc(theta) / |csc(theta)+cot(theta)|^2 d(theta)