How would you solve differential equation $\rho'(t) = 1 + \cos(t) - \rho(t)$ given particular solution in the form $B\cos(t) + C\sin(t) + D$?

Question

I am stuck at the following differential equation $$\frac{d \rho}{dt} = 1 + \cos(t) - \rho $$

Solve the non-homogenous rate equation for $\rho(t)$ given the initial condition: $\rho(t)=\rho_{0}$ at $t=0$.

Hint: seek a particular solution in the form: $$B\cos(t) + C\sin(t) + D$$ Does anyone have any ideas?

Answer 1

$$y'+y=1+\cos t$$ Solve the homogeneous equaion first: $$y'+y=0 \implies r+1=0 \implies r=-1$$ $$\implies y_h(t)=c_1e^{-t}$$ Plug the particular solution $y_p=B\cos(t) + C\sin(t) + D $ you 're given in the DE and find the constants. $$y'+y=1 +\cos t$$ $$(B\cos(t) + C\sin(t) + D )'+B\cos(t) + C\sin(t) + D =1+\cos t$$

Then the solution to the DE is: $$y(t)=y_h(t)+y_p(t)$$