Here is a sourcefree Series LRC Circuit: By applying KVL we can generate a 2ndorder Differential Equation. L di/dt + R i + 1/C ∫i dt + V_{o} = 0 Now differentiate this equation to eliminate the integral: L d²i/dt² + R di/dt + i/C = 0 Let i = Ae^{st}, therefore di/dt = si and d²i/dt² = s²i We now have the following equation: L s²i + R si + i/C = 0 → i [L s² + R s + 1/C] = 0 Since i ≠ 0, then we use L s² + R s + 1/C = 0 to find s Using the quadratic formula: s = [ b ± √(b²  4ac) ] / 2a , and letting a=L, b=R, and c=1/C We get the following equation: s = [ R ± √(R²  4L/C) ] / 2L Note there are two roots, s_{1} and s_{2}. These roots are called the natural frequencies. s_{1}, s_{2} = R/2L ± √[ (R/2L)²  1/LC ] 

For any series LRC circuit, the following equation can be used to find s_{1} and s_{2}: s_{1}, s_{2} = R/2L ± √ [ (R/2L)²  1/LC ] Note: How s_{1} and s_{2} are used is shown in another section. 
Here is a sourcefree Parallel LRC Circuit: By applying KCL we can generate a 2ndorder Differential Equation. C dv/dt + 1/L ∫v dt + I_{o} + V/R = 0 Now differentiate this equation to eliminate the integral: C d²v/dt² + v/L + 1/R dv/dt = 0 Let v(t) = Ae^{st}, therefore dv/dt = s v(t) and d²v/dt² = s² v(t) Substituting v(t), we get the following equation: C s²v + 1/R sv + v/L = 0 → v [C s² + 1/R s + 1/L] = 0 Since v(t) ≠ 0, then we use C s² + 1/R s + 1/L = 0 to find s Using the quadratic formula: s = [ b ± √(b²  4ac) ] / 2a , and letting a=C, b=1/R, and c=1/L We get the following equation: s = [ 1/R ± √(1/R²  4C/L) ] / 2C Note there are two roots, s_{1} and s_{2}. These roots are called the natural frequencies. s_{1}, s_{2} = 1/(2RC) ± √[ (1/2RC)²  1/LC ] 

For any parallel LRC circuit, the following equation can be used to find s_{1} and s_{2}: s_{1}, s_{2} = 1/(2RC) ± √[ (1/2RC)²  1/LC ] Note: How s_{1} and s_{2} are used is shown in another section. 
The roots can be both real and distinct. This occurs when the discriminant is positive.  s_{1} ≠ s_{2} 
The roots can be complex numbers. This occurs when the discriminant is negative.  s_{1} = α + jβ and s_{2} = α  jβ 
The roots can be identical. This occurs when the discriminant is zero.  s_{1} = s_{2} 
s_{1} ≠ s_{2}  V(t) = A e^{s1t} + B e^{s2t} where A and B are constants.  Overdamped Case 
s_{1} = α + jβ and s_{2} = α  jβ  V(t) = e^{αt} [ A cosβt + B sinβt ] where A and B are constants.  Underdamped Case 
s_{1} = s_{2} = s  V(t) = (A + Bt) e^{st} where A and B are constants.  Criticallydamped Case 
Here is a series LRC circuit. Therefore to find the natural/transient response, we apply the series formula: s_{1}, s_{2} = R/2L ± √ [ (R/2L)²  1/LC ] s_{1}, s_{2} = 100/2m ± √ [ (100/2m)²  1/10^{12} ] s_{1}, s_{2} = 50,000 ± √ [ (25 × 10^{8}  10^{12} ] s_{1} = 50,000 + j (3 × 10^{6}) s_{2} = 50,000  j (3 × 10^{6}) Note that the roots are Complex. 

Therefore the natural response is: V(t) = e^{50,000t} [A cos (3×10^{6})t + B sin (3×10^{6})t] (Note that A and B are constants which are found with initial conditions.) V(t) is sinusoidal, therefore we say that the natural response "rings" or resonates. For this reason, the response is Underdamped. Here is the underdamped response: → 
Here is another series LRC circuit. Therefore to find the natural/transient response, we apply the series formula: s_{1}, s_{2} = R/2L ± √ [ (R/2L)²  1/LC ] s_{1}, s_{2} = 100/(0.2m) ± √ [ (100/(0.2m))²  1/(0.01n) ] Note that m = 10^{3} and n = 10^{9} s_{1}, s_{2} = 500,000 ± √ [ (500,000)²  (100 × 10^{9}) ] s_{1}, s_{2} = 500,000 ± √ [ 15 × 10^{10} ] s_{1}, s_{2} = 500,000 ± 387,000 s_{1} = 887k and s_{2} = 113k Note that k=1000 Note that the two roots are both real, but distinct from each other. 

Therefore the natural response is: V(t) = Ae^{887kt} + Be^{113kt} (Note that A and B are constants which are found with initial conditions.) The response is Overdamped. Here is the overdamped response: → 
Here is another series LRC circuit. Therefore to find the natural/transient response, we apply the series formula: s_{1}, s_{2} = R/2L ± √ [ (R/2L)²  1/LC ] s_{1}, s_{2} = 100/0.2m ± √ [ (100/0.2m)²  1/4p ] Note p=10^{12} s_{1}, s_{2} = 500,000 ± √ [ (500,000)²  0.25 × 10^{12} ] s_{1}, s_{2} = 500,000 ± √ [ 25 × 10^{10}  0.25 × 10^{12} ] s_{1}, s_{2} = 500,000 ± √ [ 0 ] s_{1}, s_{2} = 500,000 Note that s_{1} and s_{2} are identical. Therefore this circuit is critically damped. 

Therefore the natural response is: V(t) = Ae^{500kt} + Bte^{500kt} (Note that A and B are constants which are found with initial conditions.) The response is critically damped. Here is the critically damped response: → 
Assume the following Natural Response: V(t) = Ae^{4t} + Be^{t} Also assume you are given the following information about the circuit: V(t=0) = 4 and dV/dt(t=0) = 11 Note that two Initial Conditions are needed because there are two unknowns, A and B. 
We now apply the initial conditions to find A and B: V(t=0) = 4 = Ae^{0} + Be^{0} 4 = A + B EQN 1 Using the other initial condition: dV/dt = 4Ae^{4t}  Be^{t} dV/dt(t=0) = 11 = 4Ae^{0}  Be^{0} 11 = 4A  B EQN 2 Substitute (4  A) for B in EQN 2: 11 = 4A  (4  A) 15 = 3A A = 5 therefore B = 9 The Natural Response is: V(t) = 5e^{4t} + 9e^{t} 
Assume the following Natural Response: i(t) = e^{2t}(A cos3t + B sin3t) Also assume you are given the following information about the circuit: i(t=0) = 3 and di/dt(t=0) = 6 
We now apply the initial conditions to find A and B: i(t=0) = 3 = e^{0}(A cos0 + B sin0) 3 = (A cos0) = A × 1 = A EQN 1 di/dt = 2e^{2t}(A cos3t + B sin3t) + e^{2t}(3A sin3t + 3B cos3t) di/dt(t=0) = 6 = 2e^{0}(A cos0 + B sin0) + e^{0}(3A sin0 + 3B cos0) di/dt(t=0) = 6 = 2(A) + (3B) EQN 2 Plug A=3 into EQN 2: 6 = 2(3) + 3B 12 = 3B B = 4 The Natural Response is: i(t) = e^{2t}(3 cos3t + 4 sin3t) 
Assume the following Natural Response: V(t) = (A + Bt)e^{3t} Also assume you are given the following information about the circuit: V(t=0) = 4 & dV/dt(t=0) = 8 
We now apply the initial conditions to find A and B: V(0) = 4 = (A + 0)e^{0} = A A = 4 EQN 1 dV/dt = Be^{3t}  3(A + Bt)e^{3t} dV/dt(t=0) = Be^{0}  3(A + 0)e^{0} dV/dt(t=0) = 8 = B  3A EQN 2 8 = B 3(4) 8 + 12 = B B = 20 The Natural Response is: V(t) = (4 + 20t)e^{3t} 
Assume there is a Forced Response. This will get added to the Natural Response. Assume the following Complete Response. i(t) = Ae^{4t} + Be^{t} + 20 Note that the 20 is the forced response  i.e. what is left after the transient response has died out. Also assume you are given the following information about the circuit: i(t=0) = 20 and di/dt(t=0) = 15 
We now apply the initial conditions to find A and B: i(t=0) = 20 = A + B + 20 A = B EQN 1 di/dt = 4Ae^{4t}  Be^{t} di/dt(t=0) = 4Ae^{0}  Be^{0} di/dt(t=0) = 15 = 4A  B EQN 2 15 = 4A  (A) 15 = 3A A = 5 B = +5 The Complete Response is: i(t) = 5e^{4t} + 5e^{t} + 20 