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Prove It From DE 1: $\displaystyle \displaystyle \frac{dx}{dt} + y = \sin{t}$
$\displaystyle \displaystyle y = -\frac{dx}{dt} + \sin{t}$
$\displaystyle \displaystyle \frac{dy}{dt} = -\frac{d^2x}{dt^2} + \cos{t}$.
Substituting into DE 2:
$\displaystyle \displaystyle -\frac{d^2x}{dt^2} + \cos{t} + x = \cos{t}$
$\displaystyle \displaystyle -\frac{d^2x}{dt^2} + x = 0$
$\displaystyle \displaystyle \frac{d^2x}{dt^2} - x = 0$.
Now, the characteristic equation is $\displaystyle \displaystyle m^2 - 1 = 0$
$\displaystyle \displaystyle m^2 = 1$
$\displaystyle \displaystyle m = \pm 1$.
So the solution to the homogeneous equation is $\displaystyle \displaystyle x = Ae^{t} + Be^{-t}$.
Follow a similar process to find $\displaystyle \displaystyle y$.