# Trouble 'reversing' $\frac{dv(t)}{dt}=L\,\frac{d^2i(t)}{dt^2}+R\,\frac{di(t)}{dt}+\frac{1}{C}\,i(t)$

I'm having trouble with this differential equation. I would like to reverse it, and, in particular, express it in terms of $$i(t)$$ with on the other side some linear (integro-)differential operator on $$v(t)$$ $$\frac{dv(t)}{dt}=L\,\frac{d^2i(t)}{dt^2}+R\,\frac{di(t)}{dt}+\frac{1}{C}\,i(t)\tag{1}$$

This is the common ODE of a RCL electric circuit with $$v(t),\,i(t):\mathbb R\to\mathbb R$$ and $$L,\,R,\,C\in\mathbb R_+$$

Can anyone help me with the calculation i've done or with a totally new approach. Thanks a lot.

Note: I don't want to solve it, but my aim is to 'invert it', i.e. obtain some $$i(t)=\mathcal Y[v(t)]$$ with $$Y[\,\cdot\,]$$ linear operator, if possible.

What i've tried:

Laplace transform with $$V(s)\equiv \mathcal L[v(t)]$$ and $$I(s)\equiv \mathcal L[i(t)]$$, leads to $$I(s)=\frac{s}{s^2\,L+s\,R+1/C}\,V(s)\tag{2}$$ Using Partial fraction decomposition, for suitable $$a,b \in\mathbb C$$, i can rewrite it as $$I(s)=\left(\frac{a}{s-s_1}+\frac{b}{s-s_2}\right)\,V(s)\tag{3}$$ where $$s_1,\,s_2$$ are the roots of the polynomial at denominator in eq. (2). Note that $$a+b=1$$.

Now, to come back to $$t$$ domain, i compute the Inverse Laplace transform, so that \begin{align*} i(t)&=\frac{1}{2\pi i}\,\lim_{T\to +\infty}\int_{\gamma-iT}^{\gamma+iT}I(s)\,e^{st}\,ds\\ &=\frac{1}{2\pi i}\,\lim_{T\to +\infty}\left[ a\int_{\gamma-iT}^{\gamma+iT}\frac{1}{s-s_1}\,V(s)\,e^{st}\,ds +b\int_{\gamma-iT}^{\gamma+iT}\frac{1}{s-s_2}\,V(s)\,e^{st}\,ds \right] \end{align*} Now, i operate a change of variables, so that $$\tilde s\equiv s-s_1$$ and $$\overline s\equiv s-s_2$$, then \begin{align*} i(t)&=\frac{1}{2\pi i}\,\lim_{T\to +\infty}\left[ a\,e^{s_1 t}\int_{\gamma-iT}^{\gamma+iT}\frac{1}{\tilde s}\,V(\tilde s+s_1)\,e^{\tilde st}\,d\tilde s +b\,e^{s_2 t}\int_{\gamma-iT}^{\gamma+iT}\frac{1}{\overline s}\,V(\overline s+s_2)\,e^{\overline st}\,d\overline s \right] \end{align*} By Laplace transform shifting property, i have that $$V(\tilde s+s_1)=\mathcal L[v(t)\,e^{-s_1 t}]$$ and $$V(\overline s+s_2)=\mathcal L[v(t)\,\,e^{-s_2 t}]$$, so substituting this into last equation, and further recognizing Laplace transform of an integral, i can write \begin{align*} i(t)&=\frac{1}{2\pi i}\,\lim_{T\to +\infty}\left[ a\,e^{s_1 t}\int_{\gamma-iT}^{\gamma+iT}\frac{1}{\tilde s}\,\mathcal L[v(t)\,e^{-s_1 t}]\,e^{\tilde st}\,d\tilde s +b\,e^{s_2 t}\int_{\gamma-iT}^{\gamma+iT}\frac{1}{\overline s}\,\mathcal L[v(t)\,e^{-s_2 t}]\,e^{\overline st}\,d\overline s \right]\\ &=\frac{1}{2\pi i}\,\left[ a\int_0^tv(\tau)\,d\tau +b\int_0^tv(\tau)\,d\tau \right]\\ &=\frac{1}{2\pi i}\,(a+b)\int_0^tv(\tau)\,d\tau\\ &=\frac{1}{2\pi i}\,\int_0^tv(\tau)\,d\tau \end{align*} which is clearly wrong, and i don't understand where mistake is.

Edit: Mistake is in last step, let me rewrite it \begin{align*} i(t)&=\frac{1}{2\pi i}\,\lim_{T\to +\infty}\left[ a\,e^{s_1 t}\int_{\gamma-iT}^{\gamma+iT}\frac{1}{\tilde s}\,\mathcal L[v(t)\,e^{-s_1 t}]\,e^{\tilde st}\,d\tilde s +b\,e^{s_2 t}\int_{\gamma-iT}^{\gamma+iT}\frac{1}{\overline s}\,\mathcal L[v(t)\,e^{-s_2 t}]\,e^{\overline st}\,d\overline s \right]\\ &=\frac{1}{2\pi i}\,a\,e^{s_1 t}\int_0^tv(\tau)\,e^{-s_1 \tau}\,d\tau +\frac{1}{2\pi i}\,b\,e^{s_2 t}\int_0^tv(\tau)\,e^{-s_2 \tau}\,d\tau \end{align*}

And computing $$a,b,s_1,s_2$$ concludes the 'reversing' process, as indeed some linear integro-differential operator applied on $$v(t)$$ was found. Yes, at the end the quest turned out to be somewhat illusory, as this eventually coincides with ODE solution of (1).

Still, i've some dubts on the change of variables, regarding the extremes of integration, how to treat them

• Oh yes, you are perfectly right, i cannot take that $e^{-s_2 t}$ out of $\mathcal L[\,\cdot\,]$.. Also this fact would make that step unmotivated.. Commented Sep 18, 2021 at 10:55
• Made correction, unfortunately result seems unvaried.. Commented Sep 18, 2021 at 11:09
• Once you write $i(t)=\text{known stuff}$ you have in some sense solved the equation. In any case, the integral operator $\mathcal{Y}$ you seek is an integral over the appropriate Green's function
– Sal
Commented Sep 18, 2021 at 17:18
• @Sal Unfortunately i don't know enough about green function such to solve with it Commented Sep 20, 2021 at 12:45