Shortcut solution for $f''(t)+c^2f(t)=0$? I am trying to prove that the general solution to $f''(t)+c^2f(t)=0$ is $f(t) = acos( ct) + bsin (ct)$
Could you please tell me if it is possible to solve it this way?
$$
\frac{d}{dt}(\frac{df}{dt})=-c^2f(t)
$$
$$
df* \frac{d}{dt}(\frac{df}{dt})=-c^2f(t)*df
$$
$$
d(\frac{df}{dt})^2=-c^2 f(t)df
$$
If not for the constant C resulting from integration, this would lead to the desired solution, but I think I am not doing it right somehow.
Thanks in advance for any suggestions.
 A: HINT
Here it is another way to approach it
\begin{align*}
y'' + c^{2}y = 0 & \Longleftrightarrow y''y' + c^{2}yy' = 0\\\\
& \Longleftrightarrow  (y')^{2} + c^{2}y^{2} = k^{2}\\\\
& \Longleftrightarrow y' = \pm\sqrt{k^{2} - c^{2}y^{2}}\\\\
& \Longleftrightarrow \frac{y'}{\sqrt{k^{2} - c^{2}y^{2}}} = \pm1\\\\
\end{align*}
Can you take it from here?
A: As far as "shortcuts" go: since this is a 2nd order autonomous equation, its solution set has dimension $2$. So if you can demonstrate two solutions, one not a scalar multiple of the other, then it's automatic that all solutions are linear combinations of those two.
The nature of this equation screams $\sin(cx)$ and $\cos(cx)$ if you are familiar with what happens during differentiation of trig functions and the chain rule.
So that would be one shortcut here to reach the solution set.
A: $$f''(t)+c^2f(t)=0$$
Maybe you wanted to do this ?
$$\dfrac {df'}{dt}=-c^2 f(t)$$
$$\dfrac {df'}{df}\dfrac {df}{dt}=-c^2 f(t)$$
$${df'}\dfrac {df}{dt}=-c^2 f(t)df(t)$$
$$f'{df'}=-c^2 f(t)df(t)$$
Now you can integrate.
$$f'^2=-c^2 f^2+k$$
This is separable.
A: Another way to do it.
$$f''+c^2f=0$$ Switch variables
$$-\frac {t''}{[t']^3}+c^2f=0$$ reduction of order $p=t'$
$$-\frac {p'}{p^3}+c^2f=0\implies p=\pm\frac{1}{\sqrt{k_1-c^2 f^2}}$$
$$t+k_2=\pm \int \frac {df}{\sqrt{k_1-c^2 f^2}}$$ Just finish and inverse.
A: Another way to do this is via the method of Frobenius. We seek solutions of the form $y=\sum_{k\geq 0} a_k x^k$. Differentiating and collecting like terms, we have
$$\sum_{k\geq 2} k(k-1)a_k x^{k-2}+c^2 \sum_{k\geq 0} a_k x^k = \sum_{k\geq 0}\left[(k+2)(k+1)a_{k+2}-c^2 a_k\right]x^k=0.$$
Matching term-by-term, we have $(k+2)(k+1)a_{k+2}=-c^2 a_k$ for all $k\geq 0$. This is solved as $$a_{2k}= \frac{(-c)^k}{(2k)!}a_0,\quad a_{2k+1}= \frac{(-c)^k}{(2k+1)!}a_1.$$
Hence
\begin{align}
y(x)&=a_0\left(1-\frac{(cx)^2}{2!}+\frac{(cx)^4}{4!}+\cdots\right)+a_1 \left(cx-\frac{(cx)^3}{3!}+\frac{(cx)^5}{5!}-\cdots\right)\\&=a_0 \cos(cx)+a_1 \sin(cx)
\end{align}
as expected.
