Hey,
Thank you for the response.
But, here I am only interested in to find out the term vc[t] and b[t] and as I did it manually the expected solution is in the form of = e^at[C1cos(bt)+C2sin(bt)]-d2. Where, a = kr/2klc1 and b = sqrt(kl^2c1^2-4krc1/2klc1). How do it get it in that cos and sin or in exponential format.
{b -> Function[{t}, (d2 t)/kl + C[1]],
bc -> Function[{t}, -((c1 d2 kr)/kl)],
vc -> Function[{t}, kr ((d2 t)/kl + C[1])],
vl -> Function[{t}, -d2 + kr ((d2 t)/kl + C[1])]}
{c1 -> (d2 t + kl C[1])/(d2 kr),
vin -> (d1 kl - d2 kl + d2 kr t + 2 d2 q1 t + d2 q2 t +
kl kr C[1] + 2 kl q1 C[1] + kl q2 C[1])/kl}, {d2 -> 0, vin -> d1,
C[1] -> 0}, {kr -> 0, vin -> d1 - d2,
C[1] -> -((d2 t)/kl)}, {kr -> 0, t -> 0, vin -> d1 - d2, C[1] -> 0}}
This is the result I got whenI followed your instruction it could not follow it.