y'=[(a)(cos(t)+b)]y(t)-y(t)^3, solve

Question

a & b are constants

Answer 1

y'=dy/dt=[a(cos(t)+b)]yt-yt³, so:

(1/y)dy/dt=[a(cos(t)+b)]t-t³=atcos(t)+abt-t³.

∫dy/y=∫(atcos(t)+abt-t³)dt=a∫tcos(t)dt+½abt²-¼t⁴+C where C is the constant of integration.

To solve ∫tcos(t)dt, let u=t, then du=dt and let dv=cos(t)dt, then v=sin(t).

∫tcos(t)dt=∫udv=uv-∫vdu=tsin(t)-∫sin(t)dt=tsin(t)+cos(t).

∫dy/y=ln(y)=atsin(t)+acos(t)+½abt²-¼t⁴+C, or y=Ae^{atsin(t)+acos(t)+½abt²-¼t⁴}, where A=e^C (i.e., another constant replacing C).