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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

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  • $\begingroup$ yes, Vs=u(t) and I can substitute it in the equation but then I' involves Dirac function which I'm not sure about that, In other words I'm confused about $t=0^-$ and $t=0^+$. $V_0$ is a constant and it's not related to $V_s$. Doublet function is the derivative of Dirac function, this is all I know about it. $\endgroup$ – Panda Nov 20 '16 at 21:10
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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.

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  • $\begingroup$ Wow, How a second order equation converts to a first order?! but now how to solve this? and is there general way to solve them or always I should try to eliminate them? $\endgroup$ – Panda Nov 20 '16 at 21:20

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