this d.e. is killing me! It's been a while since I took a d.e. class and was hoping for some help with a problem.
The equation is: $$L\frac{di}{dt}+Ri=20\sin(800t+25^\circ)$$
It is the phase shift that is getting me. I can get the transient solution but I cannot figure out the steady state.
The method of unknown coefficients doesn't seem to work. (I imagine because of the phase shift)
Any help would be appreciated!
On
Have you tried $$A\sin(800t+25^\circ)+B\cos (800t+25^\circ)?$$
for a particular solution?
It should work out.
Just take derivative and plug in
$$L\frac{di}{dt}+Ri=20\sin(800t+25^\circ)$$
to Find A and B.
On
$$L\frac{di}{dt}+Ri=20\sin(800t+25^\circ)$$ HINT :
If the phase shift is only that is getting you, simply make the phase shift disappear from the equation.
$\sin(a+b)=\cos(b)\sin(a)+\sin(b)\cos(a)$ $$L\frac{di}{dt}+Ri=20\cos(25^\circ)\sin(800t)+20\sin(25^\circ)\cos(800t)$$ Now, there is no phase shift in it.
Hint
$$Li'+Ri=20\sin(800t+25^\circ)$$ $$i'+\frac R Li=\frac {20}L\sin(800t+25^\circ)$$ Use integrating factor $$(ie^ {\frac R Lt})'=e^ {\frac R Lt}\frac {20}L\sin(800t+25^\circ)$$ $$ie^ {\frac R Lt}=\frac {20}L\int e^ {\frac R Lt}\sin(800t+25^\circ)dt$$ substitute $s=800t+25 \implies ds=800dt$ $$ie^ {\frac R Lt}=\frac 1 {40L}\int e^ {\frac R L(s-25)/800}\sin(s)ds$$ $$i(t)=\frac 1 {40L}e^ {-\frac R L(\frac 1{32}+t)}\int e^ {\frac {Rs} {800L}}\sin(s)ds$$ Just integrate now...