Given that ( y₁ ) is a solution to ( y'' − y' = 3sin(t) ) and ( y₂ ) is a solution to ( y'' − y' = −15e²ᵗ ), use the superposition principle to find a particular solution to the differential equation ( y'' − y' = 3sin(t) − 15e²ᵗ ).

A) l( yₚ(t) = 3sin(t) − 15e²ᵗ )
B) ( yₚ(t) = 3sin(t) + 15e²ᵗ )
C) ( yₚ(t) = 3sin(t) )
D) ( yₚ(t) = −15e²ᵗ )