y' + siny + xcosy + x = 0
The Attempt at a Solution
well, I've used Weierstrauss substitutions: siny = 2t/(1+t^2) , cosy = (1-t^2)/(1+t^2) , dy = 2dt/(1+t^2) where t=tan(x/2).
2/(1+t^2)dt + 2t/(1+t^2)dx + x(1-t^2)/(1+t^2)dx + xdx = 0
2dt + 2tdx + x(1-t^2)dx + x(1+t^2)dx = 0
2dt + (2t + x(1-t^2) + x(1+t^2))dx = 0
2dt + (2t + x - xt^2 + x + xt^2)dx = 0
2dt + (2t + 2x)dx = 0
dt + (t+x)dx = 0
well, if I multiply e^x to both sides, it'll be an exact differential and after integrating I'll have:
te^x + xe^x - e^x + C. after substituting t= tan(x/2) we obtain:
e^x(tan(y/2) + x - 1) + C.
well, this is a problem that our professor gave in the last lecture for extra score in the final exam. so I wanna be sure that I've solved it and my method is correct.
Is it correct if I substitute t=tan(x/2)? because the original problem is defined everywhere but after I use Weierstrauss substitutions then I should restrict x to the x's that are defined. (x=pi and -pi are obviously not defined now). Wouldn't that cause problems?