I've been practicing some differential equation questions but unfortunately am stuck on how to solve this one, all help is appreciated!

$C>0$ and $L>0$ At time $t=0, dI/dt= 4$

for times $t>0$, $I$ is described by the following differential equation:

$LC d^2 I/dt^2 + L dI/dt + 10 I = 0$

Show, by solving the equation, that current I will oscillate if $L<40C$ but will not oscillate if $L>40C$

I'd really really appreciate any help in how to solve this differential equation, it has me stumped given that L & C are constants but not numerically defined Thanks in advance


This is tricky because there are so many variables, but just solve it like any normal second-order linear ODE. First, find the characteristic polynomial:


Then, use the quadratic formula and solve:


As you already probably know, the general solution to the differential equation is $c_1e^{r_1t}+c_2e^{r_2t}$. Now, we could get rid of one of the constants using the fact that $\frac{dI}{dt}=4$, but I don't think we actually need to do that to answer the question.

Instead, let's just kind of analyze the roots to guess what the solutions behavior's going to be. First, there are $L$s on both top and bottom, so we can divide numerator and denominator by $\sqrt L$, which might help:

$$r_1=\frac{-\sqrt L+\sqrt{L-40C}}{2C\sqrt L},r_2=\frac{-\sqrt L-\sqrt{L-40C}}{2C\sqrt L}$$

Now, since $C$ and $L$ are positive, the denominator's always going to be positive, so we don't really need to analyze that. Next, let's look at the numerator. Because of the $\sqrt{L-40C}$, there are really two cases we need to examine here:

  • $L-40C < 0$, or $L<40C\rightarrow$ In this case, $\sqrt{L-40C}$ is imaginary, so we get two complex roots. As you already probably know, linear ODEs with complex roots leads to sinusoidal terms, meaning that the current $I$ will oscillate.
  • $L-40C > 0$, or $L>40C\rightarrow$ In this case, $\sqrt{L-40C}$ is real, but since $C > 0$, it is less than the $\sqrt{L}$ term. Therefore, the $-\sqrt{L}$ term always overpowers $+\sqrt{L-40C}$, which leads to two negative roots. Again, as you already probably know, linear ODEs with all negative roots lead to exponential decay terms, meaning that the current $I$ will tend toward $0$ instead of oscillating.

That analysis basically answers the question. Anyways, I hope that helps you understand how to analyze linear ODEs quantitatively!

Now, for the second part of the problem, since $L<40C$, $\sqrt{L-40C}$ is imaginary, so we can write the roots as a complex numbers as follows:

$$r_1=-\frac{1}{2C}+\frac{\sqrt{40C-L}}{2C\sqrt{L}}i, r_2=-\frac{1}{2C}-\frac{\sqrt{40C-L}}{2C\sqrt{L}}i$$

Now, because we have complex roots, we know that the solution is in the form of $c_1e^{at}\sin(bt)+c_2e^{at}\cos(bt)$ where $a=-\frac{1}{C}$ is the real part of the roots and $b=\frac{\sqrt{40C-L}}{2C\sqrt{L}}$ is the imaginary part of the roots. Therefore, we find the following:


Now, if $I(0)=0$, then we can plug in $t=0$ to the above equation to find the following:

$$0=c_1e^0\sin(0)+c_2e^0\cos(0)\rightarrow c_2=0$$

We can now plug in $c_2=0$ into the above equation to simplify our expression for $I(t)$:


Finally, to solve for $c_1$, we can use the fact that $I'(0)=4$. First, let's find what $I'(t)$ is:


Now, plug in $t=0$:

$$4=-\frac{c_1}{2C}e^0\sin(0)+c_1\frac{\sqrt{40C-L}}{2C\sqrt{L}}e^{0}\cos(0)\rightarrow c_1\frac{\sqrt{40C-L}}{2C\sqrt{L}}=4\rightarrow c_1=\frac{4}{\frac{\sqrt{40C-L}}{2C\sqrt{L}}}$$

Thus, our final expression for $I(t)$ is:


Now, they say $I(t)$ should be in the form $\frac 4 \omega e^{-t/(2C)}\sin(\omega t)$. Clearly, our equation for $I(t)$ matches that form, and the expression for $\omega$ is $\frac{\sqrt{40C-L}}{2C\sqrt{L}}$.

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  • $\begingroup$ This is amazing, thank you so so much, you really cleared this up!!! $\endgroup$ – Max Dec 30 '18 at 13:28
  • $\begingroup$ @Max Just so you know, if this answer was the most helpful to you, you can mark it as the accepted answer by clicking the checkmark on the left side of my answer, right below the downvote button. Thanks! $\endgroup$ – Noble Mushtak Dec 30 '18 at 13:46
  • $\begingroup$ Hi Noble, thanks I've marked it as accepted now Would you be able to help me with the second part of this practise question? I've honestly tried so hard with it, but again the solution eludes me.. and you explained the above so well: "Assuming L<40C, show that the solution has the form shown below and write down an expression for ω in terms of L & C: I(t)=4/ω. e^(-t/2c) . sinωt Thank you in advance! @noble Mushtak $\endgroup$ – Max Dec 31 '18 at 15:41
  • $\begingroup$ @Max I have answered this second part of the question by editing my answer above, so please read my new answer. Also, I am not sure if this was in the problem statement, but in order for the problem to work out, I had to assume at $t=0$, $I=0$ and $\frac{dI}{dt}=4$. $\endgroup$ – Noble Mushtak Dec 31 '18 at 17:00

This is a standard second-order form, with solution:

$$I(t) = e^{-\frac{\sqrt{-l-10} t}{\sqrt{c} \sqrt{l}}} \left(c_1 e^{\frac{2 \sqrt{-l-10} t}{\sqrt{c} \sqrt{l}}}-c_1+4\right)$$

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  • $\begingroup$ Why have the $+4$ in the end? Wouldn't the solution be the same without it because of $c_1$? Also, where are your exponents coming from? I am getting quite a different solution than you. $\endgroup$ – Noble Mushtak Dec 30 '18 at 1:48

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