To put this into some context, I was considering the general equation for damped harmonic motion:

$\ddot z +2\gamma\dot z +\omega_0^2z=0$

and specifically there limiting cases where the the damping force (to which the second term corresponds) is zero, and where the restoring force (to which the last term corresponds) is zero.

In the second case, I yield the differential equation

$\ddot z +2\gamma\dot z=0$

and I solved this in the following way:

I split $\ddot z$ into $\frac{d}{dt}\dot z$ and then separated the vairables:

$\int\frac{1}{\dot z}d\dot z =\int-2\gamma dt$

which gives the exponential solution expected.

Now I am considering the case where there is simple harmonic motion with no damping, i.e. the case that gives rise to the differential equation:

$\ddot x +\omega_0^2 x=0$

I can't figre out- is there a way to solve this without just assuming an exponential solution and finding the constants in the solution? For example, in the equation above I was able to solve it by separating the variables. I'm not so sure this would work here where we have a second derivative of x and x itself, but i'm not sure.

Also, I would be grateful if anyone could recommend an online resource or book that explains how to solve different types of integrals and differential equations well. I have been trying to study this for some time but haven't come across anything that seems very complete, and perhaps something that has some worked examples?

  • 2
    $\begingroup$ $\ddot x +\omega_0^2 x=0\implies 2(\dot x \ddot x +\omega_0^2 x \dot x) = 0 \implies \frac{d}{dx}(\dot x^2 +\omega_0^2 x^2)$ is constant, is a good first step. $\endgroup$
    – πr8
    Commented Jan 1, 2017 at 16:46
  • 1
    $\begingroup$ Laplace transforming both sides is also an option $\endgroup$
    – GuPe
    Commented Jan 1, 2017 at 16:54
  • 3
    $\begingroup$ Out of curiosity: why are you looking for an alternative method? I'm not saying you're wrong in doing so, quite the oposite - it can be tremendously informative, but the "standard" mathod is perfectly fine and gives you a general solution. I'm getting a vibe from your post that you feel like you're missing on some solutions by "assuming" the exponential form of it, but you're not. It's an easy excercise to prove that every solution is of this form. $\endgroup$ Commented Jan 1, 2017 at 17:12
  • $\begingroup$ What Hirek said. Plus, the general, standard, method allows you to treat in one strike all the subcases you are making (first $\gamma=0$, then $\omega_0=0$). $\endgroup$
    – Did
    Commented Jan 1, 2017 at 17:16
  • $\begingroup$ @HirekKubica One argument might be that this method is limited (sort of) to linear, constant-coefficient equations. It's also the case (in my experience) that there's a pretty wide variety of first-order ODEs (linear and beyond) and an accompanying range of techniques for getting at their solutions. Moving to higher-order ODEs, it can arguably seem a little artificial to start with these ansatz-based techniques, which can seem to assume too much to start with. The somewhat disappointing corollary is that we often can't do much better, and have to be satisfied with, e.g., existence results. $\endgroup$
    – πr8
    Commented Jan 1, 2017 at 19:10

5 Answers 5


One, not-entirely-standard solution (though it does generalise, to some extent):

Write your equation as $\frac{\ddot x}{x}=-\omega^2$, and let $u=\frac{\dot x}{x}$.

Then $\dot u=\frac{\ddot x}{x}-\left(\frac{\dot x}{x}\right)^2\implies \dot u+u^2+\omega^2=0\implies\frac{\omega\dot u}{u^2+\omega^2}+\omega=0$.

From this we see:

$$\dot{\left(\arctan\frac{u}{\omega}\right)}+\omega=0\implies u=-\omega\tan(\omega t+c)$$

for some $c$. Thus

$$\frac{\dot x}{x}=-\omega\tan(\omega t+c)\implies \dot{(\log x)}=\dot{(\log \cos (\omega t+c))}$$

and we deduce that $x=R\cos(\omega t+c)$ for some $R,c$.

  • 2
    $\begingroup$ Just a remark: writing dot for time derivative over long expressions looks kind of unsettling. $\endgroup$ Commented Jan 1, 2017 at 17:03
  • 1
    $\begingroup$ @HirekKubica I actually tend to agree with you there; was mostly for consistency with OP's post but certainly not something I'd generally go for. $\endgroup$
    – πr8
    Commented Jan 1, 2017 at 17:10
  • $\begingroup$ Thank you for your reply! However I am having a bit of difficulty understanding how you got $from the second step to the third step in the third line? $\endgroup$
    – Meep
    Commented Jan 1, 2017 at 20:52
  • $\begingroup$ (assuming I understand you correctly) it's to do with that $\frac{d}{dx}\log f(x)=\frac{f'(x)}{f(x)}$, and using this on both sides of the equation (noting that $\frac{d}{dx}\log \cos x = -\tan x$) $\endgroup$
    – πr8
    Commented Jan 1, 2017 at 20:55

Since your second order linear ordinary differential equation has constant coefficients, you can generalise the solution to your equation (below) for any values of $\gamma$ and $\omega$ without any limitations by using the concept of the characteristic polynomial.

$$\frac{d^2 z}{dt^2}+2\gamma\frac{dz}{dt}+\omega^2 z=0$$

We will assume the form of the solution to be $z(t)=e^{\lambda t}$

We will substitute this into our differential equation to obtain:

$$\lambda^2 e^{\lambda t}+2\gamma\lambda e^{\lambda t}+\omega^2 e^{\lambda t}=0$$

$$e^{\lambda t}(\lambda^2+2\gamma\lambda+\omega^2)=0$$

We deduce that $e^{\lambda t}$ will only equal zero for non-zero values of $t$ if $\lambda \to -\infty$. Hence, this is a trivial solution.

Thus, we must only consider the solution to $\lambda$ on $\lambda^2+2\gamma\lambda+\omega^2=0$ (Our characteristic polynomial).

We solve this using the quadratic formula. Therefore, the two roots of the characteristic polynomial will be:

$$\lambda=\frac{-2\gamma \pm \sqrt{4\gamma^2-4\omega^2}}{2}=-\gamma \pm \sqrt{\gamma^2-\omega^2}$$

We denote $\lambda_1$ and $\lambda_2$ to be the roots of the above equation.

We notice that depending on the values of the constants $\gamma$ and $\omega$, the solutions to $\lambda$ will be different, and so hence our general solutions will be significantly different.

We separate each case:

  1. If both roots are real and distinct (i.e. $\lambda_1 \neq \lambda_2$)
  2. If both roots are complex.
  3. Real repeated roots (i.e. $\lambda_1=\lambda_2$)

Case 1

Since the particlular solutions are $z_1(t)=e^{\lambda_1 t}$ and $z_2(t)=e^{\lambda_2 t}$, they will yield the general solution when substituted to $z(t)=k_1 z_1(t) + k_2 z_2(t)$.

$$z(t)=k_1 e^{\lambda_1 t}+k_2 e^{\lambda_2 t}$$

Case 2

We denote a complex number by $\lambda=a+bi$.

Since $\lambda_{1,2}$ will both be complex, using $e^{i \theta}=\cos{\theta}+i\sin{\theta}$, we obtain the general solution:

$$z(t)=k_1 e^{at} \cos{(bt)} + k_2 e^{at} \sin{(bt)}$$

Case 3

$$z(t)=k_1 e^{\lambda t}+k_2 t e^{\lambda t}$$


Hence, for the solution to the differential equation you wanted to find, $\ddot{z}+\omega z=0$, let $\gamma=0$. Since both solutions to $\lambda$ are purely imaginary (i.e. Complex with $a=0$), the form of the solution will be Case 2. The general solution is hence:

$$z(t)=k_1 \cos{(\omega t)} + k_2 \sin{(\omega t)}$$

You can substitute your initial conditions into the equation to obtain the values of the arbitrary constants $k_1$ and $k_2$ for simple harmonic motion as:

$$z(t)=z_0 \cos{(\omega t)} + \frac{v_0}{\omega} \sin{(\omega t)}$$

Which can be written in the form:

$$z(t)=A (\cos{\omega t+\varphi})$$

For more information, see:

Real, Distinct Roots, Complex Roots and Repeated Roots

  • $\begingroup$ Thank you for your reply! This is a really wonderful answer but unfortunately I know about this general solution and I was just asking my question in this context- I was more thinking of how it could be possible to solve the differential equation with $\ddot x$ and $x$ only without just assuming the exponential solution. But I think this answer should definitely be put somewhere for these types of questions! $\endgroup$
    – Meep
    Commented Jan 1, 2017 at 20:56

Let's first make clear that we are not assuming that the solution for this differential equation has exponential form. This equation belongs to a class of differential equations whose have simple solutions and indeed have exponential form on a more subtle sense. Now we will find out what class of differential equations are we talking about by defining a new variable.

Let $ v = \dot z $. Then $ \dot v = \ddot z = - 2\gamma\dot z - \omega_0^2z $.

So we can see that this equation is actually is a system of homogenous linear differential equations. Namely $$ \begin{cases} \dot z = v \\ \dot v = - 2\gamma v - \omega_0^2z \end{cases} $$

Which can be written in matrix form by setting $ x = \begin{bmatrix} z \\ v \end{bmatrix} $ and $ A = \begin{bmatrix} 0 & 1 \\ - \omega_0^2 & - 2\gamma \end{bmatrix} $.

Using this notation the system simply becomes $$ \dot x = Ax $$

This equation doesn't remember you anything? You should be familiar with this equation when $ A \in \mathbb{R} $ and its solution $ x(t) = ke^{At} $ where $k$ is some real constant. It is from here that arises the motivation for matrix exponentiation. Consider the following question:

It is possible to extend the definition exponentiation from real numbers to real square matrices such that it preserves all its properties including the function $ x(t) = ke^{At} $ being the solution of the (now matrix) equation $ \dot x = Ax $ ?

The answer is yes and it is done naturally by just subtituting real numbers by real matrices in the power series definition of the exponential function. That is

$$ e^A := \sum _{ k=0 }^{ \infty }{ \frac { 1 }{ k! } { A }^{ k } } $$

Even more astonishing is that this definition even works for complex matrices!

Now back to the particular case $ \ddot x +\omega_0^2 x=0 $, the reason its real solution is written in terms of trigonometric functions is that its correspoding matrix has complex eigenvalues and Euler's formula $$ e^{iz} = cos(z) + i sin(z) $$ plays a role here.

By using the following two propositions which can be proven true and Euler's Formula we derive the general real solution from the general complex solution.

Proposition 1. The general complex solution of the linear system $ \dot x = Ax $ where $A$ is a $n$ x $n$ real matrix with $n$ distinct (possible complex) eigenvalues $\lambda_1, ..., \lambda_n$ associated with eigenvectors $v_1, ..., v_n$ is given by $$ x(t) = \sum _{ i=1 }^{ n }{ c_i e^{\lambda_i t} v_i } $$ where $c_i$ are complex constants. Note that this is complex exponentiation.

Proposition 2. The general real solution of the linear system $ \dot x = Ax $ where $A$ is a $n$ x $n$ real matrix with $n$ distinct (possible complex) eigenvalues $\lambda_1, ..., \lambda_n$ associated with eigenvectors $v_1, ..., v_n$ is a linear combination of real and imaginary parts of complex solutions. That is, if $x(t)$ is a complex solution, then $y(t) = k_1 Re(x(t)) + k_2 Im(x(t))$ is a real solution, where $k_1$ and $k_2$ are real constants.

(The details are rough but this is supposed to be just a glimpse of the theory behind)

Solution. Writing the equation associated matrix $ A = \begin{bmatrix} 0 & 1 \\ - \omega_0^2 & 0 \end{bmatrix} $ we find its eigenvalues $ \lambda_1 = -i w_0 $ and $ \lambda_2 = i w_0 $ with corresponding eigenvectors $v_1$ and $v_2$. Now we apply (Prop. 1) to get the complex general solution, followed by Euler's formula to get complex linear combinations of trigonometric functions and finnaly (Prop. 2) with some algebra to arrive at the general real solution given by: $$ x(t) = \begin{bmatrix} z(t) \\ v(t) \end{bmatrix} = \begin{bmatrix} k_1 cos(\omega_0 t) + k_2 sin(\omega_0 t) \\ - k_1 sin(\omega_0 t) + k_2 cos(\omega_0 t) \end{bmatrix} $$

You have to go through a bit of theory but after it is done it gets much more easier to find the general solution for this class of differential equations.

I hope I brought up some insight of a natural way to dealing with this kind of equations. You can find the details, worked examples and more about in the great book entitled Differential Equations, Dynamical Systems, and an Introduction to Chaos by Hirsch, Smale and Devaney.


This is quite the old question, but nobody has given an alternate method for solving the general case, I might as well chip in if anybody sees it.

Starting from the ODE \begin{align} \ddot z+2\gamma \dot z+\omega_0^2z=0, \end{align} let $x=\alpha z+\dot z$, \begin{align} \dot x+\frac{\omega_0^2}{\alpha}x+\left(2\gamma-\alpha-\frac{\omega_0^2}{\alpha}\right)\dot z=0. \end{align} Choosing $\alpha$ to be one of the solutions to the quadratic equation \begin{align} 2\gamma-\alpha-\frac{\omega_0^2}{\alpha}=0\longrightarrow \alpha=\gamma \pm\sqrt{\gamma^2-\omega_0^2}, \end{align} gives that \begin{align} \dot x+\frac{\omega_0^2}{\alpha}x&=0,\ \ \ \ \frac{\mathrm d}{\mathrm dt}\ln(x)=-\frac{\omega_0^2}{\alpha},\\ x(t)&=k_1\exp\left(-\frac{\omega_0^2}{\alpha}t\right). \end{align} Then to find $z(t)$, \begin{align} \dot z+\alpha z=k_1\exp\left(-\frac{\omega_0^2}{\alpha}t\right)&, \ \ \ \ \frac{\mathrm d}{\mathrm dt}\left(\exp(\alpha t)z\right)=k_1\exp\left((\alpha-\frac{\omega_0^2}{\alpha})t\right),\\ z(t)=k_1&\exp\left(-\frac{\omega_0^2}{\alpha} t\right)+k_2\exp\left(-\alpha t\right). \end{align} If we take a closer look at the factors in the exponents, \begin{align} -\frac{\omega_0^2}{\alpha}=\alpha-2\gamma=-\gamma\pm\sqrt{\gamma^2-\omega_0^2},\\ -\alpha=-\gamma\mp\sqrt{\gamma^2-\omega_0^2}, \end{align} we see we get the same as the other answers have given. The final answer is then \begin{align} z(t)=k_1\exp\left[\left(-\gamma+\sqrt{\gamma^2-\omega_0^2}\right)t\right]+k_2\exp\left[\left(-\gamma-\sqrt{\gamma^2-\omega_0^2}\right)t\right]. \end{align}


General case...


$$ \ddot{z} + 2 \gamma \dot{z} + \omega_0^2 z = 0. $$


$$ \frac{d^2 z}{dt^2} + 2 \gamma \frac{d z}{dt} + \omega_0^2 z = f(t). $$


$$ \exp\left(-\lambda_+ t \right) \frac{d}{dt} \exp\left(\lambda_+ t \right) \exp\left(-\lambda_- t \right) \frac{d}{dt} \exp\left(\lambda_- t \right) z = f(t), $$


$$ \lambda_\pm = \gamma \pm \sqrt{\gamma^2 - \omega_0^2}. $$

So the solution can be written as

$$ z = \exp\left(-\lambda_- t \right) \int dt \exp\left(+\lambda_- t \right) \exp\left(-\lambda_+ t \right) \int dt \exp\left(+\lambda_+ t \right) f(t). $$

The case

$$ \ddot{z} + 2 \gamma \dot{z} + \omega_0^2 z = 0. $$

means $f(t)=0$ so

$$ z = \exp\left(-\lambda_- t \right) \int dt \exp\left(+\lambda_- t \right) \exp\left(-\lambda_+ t \right) \int dt \exp\left(+\lambda_+ t \right) f(t) $$

$$ ... = c_+ \exp\left(\lambda_+ t \right) + c_- \exp\left(\lambda_- t \right) $$

Special case


$$ \ddot{z} + \omega^2 z = 0. $$


$$ 2 \ddot{z}\dot{z} + 2 \omega^2 z \dot{z} = 0. $$


$$ \frac{d}{dt} \left[ \dot{z}^2 + \omega^2 z^2 \right] = 0 $$

Thus $$ \dot{z}^2 + \omega^2 z^2 = K^2 $$


$$ \frac{dz}{dt} = \sqrt{K^2 - \omega^2 z^2 } $$


$$ t = \int \frac{dz}{\sqrt{K^2 - \omega^2 z^2 }} $$


$$ t = \frac{1}{\omega}\sin^{-1} \left( \frac{z\omega}{K} \right) + t_0 $$


$$ z = K \sin\left( \omega [ t - t_0 ] \right) $$


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