Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the DGL: $y''= -k^{2}y$

I tried solving this with the quadratic form attempt: $y_{1,2} = \frac{-k^{2}}{2} \pm \frac{\sqrt{k^{4}-4}}{2}$

A good solution attempt is $A\cos(bx)+B\sin(bx) $, I know that. How does one reach there without knowing this solution attempt?

share|cite|improve this question
up vote 4 down vote accepted

The trigonometric functions $\cos(x)$ and $\sin(x)$ both satisfy $y'' = -y$.

It is straightforward to show that both, $\cos(kx)$ and $\sin(kx)$ satisfy $y'' = -k^2y$ and your result follows by the superposition principle, since the DE is linear.

share|cite|improve this answer

In general, for a contant-coefficient homogeneous linear differential equation $P(D) y = 0$, where $D = \frac{d}{dx}$ and $P$ is a polynomial of degree $n$ with $n$ distinct roots $r_1, \ldots, r_n$, the general solution is $y(x) = \sum_{j=1}^n c_j e^{r_j x}$ where $c_1, \ldots, c_n$ are arbitrary constants. In your case, assuming $k \ne 0$, $P(t) = t^2 + k^2$ has the two roots $\pm k i$, so the general solution is $y(x) = c_1 e^{ikx} + c_2 e^{-ikx}$. Since $e^{ikx} = \cos(kx) + i \sin(kx)$, an alternate form of the general solution is $y(x) = A \cos(k x) + B \sin(k x)$, where $A = c_1 + c_2$ and $B = i c_1 - i c_2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.