I want to know the right way for solving differential equations which involve Dirac function. I'm not expert in mathematics, so please answer as simple as possible. suppose I want to solve this example: (This is the equation of a circuit which I want to find it's step response)
$I''(t)+3I'(t)+2I(t)=V_s''(t)+2V'_s(t)+2V_s(t)$
$I'(0)=\frac{d(V_s)}{dt}-V_s(0)-I_0+2V_0$
$I(0)=I_0-V_0+V_s(0)$
and $V_s$ is $u(t)$.
my solution:
First, I solve characteristic equation and find homogeneous answer which is:
$I_h(t)=(K_1 e^{-2t}+K_2e^{-t})$ for $t>0$
then we need a private solution for this which holds below equation:
$2*C=2$ and so $C=1$
(I supposed the private answer is a constant and also it holds eqaution for $t>0$ ) but my answer is wrong and the constant is 0.5. Where did I do wrong?
and another question is if we write this equation for $t>0$ then what's the role of $V'_s $ and $V''_s$ if $V_s$ is step function?
thanks in advance
Define $y(t)=I(t)-V_s(t)$. Then $$ y''+3y'+2y=V_s''+2V_s'+2V_s-(V_s''+3V_s'+2V_s)=-V_s' $$ which already looks like a less scary equation. Now multiply with $e^t$ and integrate, $$ (e^t(y'+2y))'=-e^tV_s'=-(e^tV_s)'+e^tV_s \\~\\ \implies e^t(y'+2y)=-e^tV_s+\int e^tV_s\,dt $$ where all Delta functions are now eliminated.
continuation: Now multiply again with $e^t$ to capture the second characteristic value to get $$ (e^{2t}y)'=-e^{2t}V_s+e^t\int e^τV_s(τ)\,dτ=\left(e^t\int e^τV_s(τ)\,dτ\right)'-2e^{2t}V_s \\~\\ \implies e^{2t}y=e^t\int e^τV_s(τ)\,dτ-2\int e^{2τ}V_s(τ)\,dτ \\~\\ \text{ or } y(t)=e^{-t}\int e^τV_s(τ)\,dτ-2e^{-2t}\int e^{2τ}V_s(τ)\,dτ $$ Of course this is the same as performing variation of constants and some partial integration on the top equation.