(a) For part (a) do I just substitute the values and that's it? substituting the given values of L, R , C and V , second ode : 0.4q''+0+10 q=110 sin(wt) q''+25q=275 sin(wt) GS(H) : qh(t)=A cos (5t)+B sin (5t) Given q'(0)= 0 -5A sin(0) +5B cos(0)=0 , So B=0 q(0)=1 A cos (0)+B sin (0) =1 so A= 1 GS(H) = qh(t)= -5 sin(5t) and Now to find particular solution. Since R(t) = sin (wt) where w is element of all real numbers so Try qp(t) = a. t sin(wt) +b.t cos (wt) Substitute into o.d.e : q''+25q=275 sin(wt) [(2w.a cos (wt) -2w.b sin(wt) +w2 a.t sin (wt)+ b.t cos (wt)] +25 [(a.t sin(wt) + b.t cos (wt)]=275 sin (wt) cos (wt) . ( 2wa +25bt -w2.bt) + sin (wt) . ( -2wb+25at-w2at) = 275 sin (wt) Equate coefficient : ( 2wa +25bt -w2.bt) =0 -2wb+25at-w2at=275 Please help me out to find the GS(IH) with step-by step workings