zero state responce on second order question..

here is the question
**broken link removed**

from kvl
[latex]
v_s(t)-I_LR-V_L-V_c=0
[/latex]
[latex]I_c=I_L\\[/latex]
[latex]\dot{v_c}=\frac{I_c}{c}=\frac{I_L}{c}\\[/latex]
[latex]\int \frac{I_c}{c}=v_c[/latex]

[latex]
v_s(t)-I_LR-V_L-\int \frac{I_L}{c}=0
[/latex]
[latex]
\delta (t)=I_LR+V_L+\int \frac{I_L}{c}
[/latex]

and i got to this equation
[latex]
L\ddot{I_L}+\dot{I_L}R+\int \frac{I_L}{c}=\delta (t)
[/latex]

upon what theoretical basis they say the it equals
[latex]
\ddot{I_L}+\dot{I_L}+{I_L}=0
[/latex]

You start off with this equation:



This is rather difficult to work with, so you transform it into a standard second order differential equation by taking the derivative of both sides. The delta equation represents a physical impossibility: a pulse of infinite magnitude and zero duration. Its derivative is undefined. It is useful for theoretical analysis.

By setting the equation equal to zero, what you're looking for is the transient response, not the steady state response. Since the delta equation doesn't have a steady state response (which is one of the things that makes it so useful, especially for Fourier transforms of sampled data) the solution is purely transient.