Here is how I would do this problem:
The differential equation is
Ldtdi+Ri=Vssin(ωt). The associated homogeneous equation is
Ldtdi+Ri=0 or
Lidi=−dt. Integrating both sides,
Lln(i)=−t+C or
i(t)=C′e−t/L.
To find a solution to the entire equation, we look for a solution of the form
i(t)=Acos(ωt)+Bsin(ωt) so that
dtdi=−Aωsin(ωt)+Bsin(ωt) and the equation becomes
−ALωsin(ωt)+BLωcos(ωt)+RAcos(ωt)+RBsin(ωt)=(RB−ALω)sin(ωt)+(BLω+RA)cos(ωt)=Vssin(ωt).
Your solutions don't look right to me but to answer your question, you need to use the trig identity
cos(A+B)=sin(A)cos(B)+cos(A)sin(B). Then also
cos(A−B)=sin(A)cos(B)−cos(A)sin(B) because sine is an "odd" function while cosine is an "even" function. Adding the two equations,
2sin(A)cos(B)=cos(A+B)+cos(A−B).
Here is how I would do this problem:
Your equation is
Ldtdi+Ri=Vssin(ωt). The Associated homogeneous equation is
dtdi=−Ri which can be written as
idi=−LRdt. Integrating both sides,
ln(i(t))=−LRt+C or
i(t)=C′e−LRt.
To find a solution to the entire equation, look for a solution of the form
i(t)=Asin(ωt)+Bcos(ωt). Then
dtdi=−Aωcos(ωt)+Bωsin(ωt) and the equation becomes
−ALωcos(ωt)+BLωsin(ωt)+ARsin(ωt)+BRcos(ωt)=(BR−AL)cos(ωt)+(BL+AR)sin(ωt)=Vssin(ωt).
In order that this be true for all t we must have the coefficients of
sin(ωt) and of
cos(ωt) the same for both sides:
RB−ALω=Vs and
BL+RA=0. To eliminate A, multiply the first equation by R, the second equation by
Lω and add:
R2B−RAL=RVs added to
BL2ω+RAL=0 is
(R2+L2ω)B=RVs so
B=R2+L2ωRVs. Then
BL+RA=0 becomes
RA=−BL=−R2+L2ωRLVs and then
A=−R2+L2ωLVs.
So the general solution to the differential equation is
i(t)=−R2+L2ωLVscos(ωt)+R2+L2ωRVssin(ωt)+C′e−t/L.
If we also have the initial condition that i(0)= 0 then
i(o)=−R2+L2ωLVs+C′=0 so that
C′=R2+L2ωLVs and the solution to the initial value problem is
i(t)=−R2+L2ωLVscos(ωt)+R2+L2ωRVssin(ωt)+(R2+L2ωLVs)e−t/L.