No, you don't need to do two equations.
Have you evaluated the homogeneous equation's solution yet?
For "linear differential equations with constant coefficients" if the "right hand side" functions are those that we expect as solutions to a homogeneous equation, , , , powers of t, and products of those, then you can expect such functions as solutions of the equation.
Here, the right side is 7sin(2t)+ 5t cos(2t). So try a solution of the form y= Asin(2t)+ Bcos(2t)+ Ctsin(2t)+ Dtcos(2t).
Then y'= 2A cos(2t)- 2B sin(2t)+ Ct cos(2t)+ C sin(2t)- Dt sin(2t)+ D cos(2t)= (Ct+ 2A+ D)cos(2t)+ (C-2B- Dt) sin(2t) and
y''= C cos(2t)- 2(Ct+ 2A+ D)sin(2t)- D sin(2t)+ 2(C- 2B- Dt) cos(2t).
Put those into the differential equations to get equations to solve for A, B, C, and D.