Please see attached document.
1. The ODE is given as a first-order linear ODE. Here we have:
2x y' - ln(x^2) = 0
You have to use separation of variables to solve this ODE:
2x * y' = ln(x^2)
=> 2x * (dy/dx) = ln(x^2)
==> 2x*dy = ln(x^2) * dx
==> dy = ln(x^2)/2x * dx
We must now integrate both sides:
y = Integral[ln(x^2)/2x] *dx
This is not a complicated integral, since this is an ODE, I assume you know calculus, it's just integration by parts, twice:
You get, y = (1/8)*(ln(x^2))^2 + C, where C is some constant.
This is the general solution to the equation above.
2. Here, I let t = theta.
y' cos(t) = sin(t) + 2.5
Re-writing y' = dy/dt:
(dy/dt)cos(t) = sin(t) + 2.5
=> dy (cos(t)) = [sin(t) + 2.5]*dt
==> dy = [sin(t) + 2.5]/(cos(t)) * dt
Integrate both sides, by simplifying that sin[t]/cos[t] = tan[t], and 2.5/cos[t] = 2.5*sec[t]
==> y = Integral[tan[t] + 2.5*sec[t]] ...
The method of integrating factors and method of separation of variables are encompassed.