This might be a stupid question, but I have only been taught to solve 1st order 1st degree differential equations, so this one is a little hard for me.

Basically I am trying to find the current in a RLC (Resistance Inductor Capacitor) circuit as a function of time. I got the differential equation as a function of charge $q$ and it's successive derivatives with time $t$. The equation is:


I looked up a bit and found out that this was a second order differential equations, and that the solution would be a exponentially decreasing sine wave function. However I don't know how to obtain the solution, so I'd like to ask for help here.


4 Answers 4


It's been a while since I've had to explain the process of solving second order ODE's, so if anyone finds a mistake in this please do let me know

In order to solve this differential equation you would have to learn how to solve Second-Order Differential equations in general. The equation you have provided is known as a Second-Order Inhomogenous Linear Ordinary Differential Equation with Constant Coefficients.

These are of the form (Note that $q$ in this equation ONLY is not the same as your charge function q):


Second-Order: Involves the second derivative of $y$

Linear: It is of the form $p_0(x)\frac{d^ny}{dy^n}+p_1(x)\frac{d^{n-1}y}{dy^{n-1}}+p_2(x)\frac{d^{n-2}y}{dx^{n-2}}+...+p_ny(x)=r(x)$

Inhomogenous: $r(x) \neq 0$

Constant Coefficients: $p$ and $q$ are constants.

Your equation in this form is:

$$-Lq''+Rq'+\frac{q}{C}=-V\Rightarrow q''-\frac{R}{L}q'-\frac{1}{LC}q=\frac{V}{L}$$

To solve this we must do the proceed with the following steps:

Step 1: Finding the complementary solution

We start by solving the equivalent homogenous problem $y''+py'+qy=0$

Setting $V=0$

$$ q''-\frac{R}{L}q'-\frac{1}{LC}q=0$$

Now we need to find what is known as an auxillary equation. We consider that $q$ is of the form $e^{mt}$.


If we plug these in and rearrange, we get:


Since $e^{mt}\neq0$, then $(m^2-\frac{R}{L}m-\frac{1}{LC})=0$

This is a quadratic equation with solutions:

$$m_{1,2}=\frac{\frac{R}{L}\pm\sqrt{\frac{R^2}{L^2}-\frac{4}{LC}}}{2}\Rightarrow m_1=\frac{\frac{R}{L}+\sqrt{\frac{R^2}{L^2}-\frac{4}{LC}}}{2},m_2=\frac{\frac{R}{L}-\sqrt{\frac{R^2}{L^2}-\frac{4}{LC}}}{2}$$

The discriminant of the roots $\Delta = \frac{R^2}{L^2}-\frac{4}{LC}$ causes three different cases to arise:

Case 1:$\Delta>0$, which tells us we have two distinct real roots, and in this case we get two linearly independent solutions:


which gives us the complementary solution:


Case 2:$\Delta=0$, which tells us we have a double repeated root ($m_1=m_2=m$), and in this case we get two linearly independent solutions:


which gives us the complementary solution:


Case 3:$\Delta<0$, which tells us we have a two distinct complex roots and in this case we get two linearly independent solutions:

$$q_1=e^{(a+ib)t}=e^{at}(\cos bt+i\sin bt), q_2 = e^{(a-ib)t}=e^{at}(\cos bt-i\sin bt)$$

which gives us the complementary solution:

$$q_c(t)=e^{at}[C_1(\cos bt+ i\sin bt)+C_2(\cos bt - i \sin bt)]$$

Step 2: We find the particular solution (or particular integral), and this usually is a trial-and-error approach where your decision depends on the form that $r(x)$ comes in. In our case $r(x)=V$, which is just a constant function of $t$. So we will make the guess that $q_p=A$, another constant function of $t$.

Plugging this in we get: $$-\frac{A}{LC}=\frac{V}{L}\Rightarrow A=-CV$$ And if my knowledge of Electrical Physics is correct, I believe that $CV=Q$



Step 3: Now that we have our complementary and particular solutions, all we need to do is add them together to get our general solution. Hence for cases $1$ and $3$,

$$q(t)=q_c+q_p\Rightarrow q(t)=C_1e^{m_1t}+C_2e^{m_2t}-Q$$

or for case $2$,

$$q(t)=q_c+q_p\Rightarrow q(t)=C_1e^{mt}+C_2te^{mt}-Q$$

The constants $C_1,C_2$ and $Q$ can only be determined by knowing values of $q(t)$ at different times $t$, which brings two new problems in differential equations known as Initial-Value Problems and Boundary-Value Problems.

I believe in your specific equation the electrical circuit is undamped, which means $\Delta<0$ and that is how the "exponentially decreasing sine wave function arises".

Hope this gives you some insight on how to solve Second-Order ODE's.

  • 1
    $\begingroup$ Well done ! $\to +1$ $\endgroup$ Aug 18, 2017 at 18:33

$$V+\frac{q}{C}+R\frac{dq}{dt}-L\frac{d^2q}{dt^2}=0 \\\div L \\\frac{d^2q}{dt^2}-\frac{R}{L}\frac{dq}{dt}-\frac{q}{LC}=\frac{V}{L} \to\\q''+aq'+bq=c \\$$first you must solve for $q''+aq'+bq=0$ to find $q_h$ then apply R.r.s to find $q_p$ then $$q(t)=q_h+q_p$$


It is sure that, if you did not learn yet about second order differential equation, it would not be easy but le me try.

The first thing we could do here is to first set $q=z-CV$ to get $$C L z''-C R z'-z=0$$ which is slightly simpler. Now, (be sure that I am here using shortcuts) assume that $z=A e^{B t}$ (because constant coefficients plus the fact that any derivative of an exponential is an exponential) and replace to get $$A e^{B t} \left( C LB^2- C R B -1\right)=0\implies C LB^2- C R B -1=0$$ which is a quadratic equation in $B$.

You will find two roots which, more than likely, would be complex. Then, $z$ would be something like $$z=A_1 e^{B_1t}+A_2 e^{B_2t}$$ and you will need to extract in the exponentials the real part and the imaginary part, the latest leading to sine and cosine.

I hope and wish this helps.


You can first assume that the solution is in the form of exponential function, and start from here: $i(t) = Ke^{st}$ You will see that the solution is some sum of this form.

The actual solution is dependent upon damping ratio. If damping ratio is greater than 1, $i(t)=K_1 e^{s_1 t}+K_2 e ^ {s_2 t}$

If damping ratio is equal to 1, $ i(t)=K_1 e^{st} +K_2te^{st} $

If damping ratio is smaller than 1, $ i(t)=e^{-αt} (A_1 cosw_d t+A_2 sinw_d t) $

There is a 10 min video on it. It shows how to drive the actual equation and how to solve them.


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