Solving $y''-k^2y=0$ without substituting $e^{kx}$ As the title says I am trying to solve $y''-k^2y=0$. The method that I want to use is to assume $y'=p$ which gives us $y''=p\frac{dp}{dy}$. 
Substituting above values in original equation gives me $p\frac{dp}{dy}-k^2y=0$ which further reduces to $\frac{dy}{dx}=\sqrt{k^2y^2+c}$. On trying to solve this differential equation I am not reaching any close to the expected answer which should be summation of two exponential terms. 
 A: $$\frac{dy}{dx}=\sqrt{k^2y^2+c}$$
with $c=k^2a$
$$\frac{dy}{dx}=|k|\sqrt{y^2+a}$$
Substitute $y=\sqrt a\sinh(t)$
$$\frac{dy}{dx}=|k|\sqrt{a\sinh^2(t)+a}=|k|\sqrt{a\cosh^2(t)}$$
$$\frac {\sqrt{a}\cosh(t)}{\cosh(t)}dt=|k|\sqrt{a}dx$$
$$\int dt=|k|\int dx$$
$$y(x)=\sqrt a \sinh(|k|x+K)$$
Using Euler's formula
$$y(x)=c_1e^{kx}+c_2e^{-kx}$$

You can also use the fact that
$$y''-k^2y=0$$
$$\frac {y''}y=k^2$$
This equation becomes a separable diff equation of first order
$$ \implies z'+z^2=k^2$$
$$  \int \frac {dz}{k^2-z^2}=\int dx$$
$$ \int \frac {dz}{k^2-z^2}=x+K_1$$
where $z=\frac {y'}y$
A: Hint: try hyperbolic substitution $$y=a\sinh u$$or$$y=a\cosh u$$ for some proper constant $a$
A: Multiplying by $y'$ we get
$$
y'y''-k^2 y y' = \frac 12 \frac{d}{dt}(y')^2-k^2\left(\frac 12\frac{d}{dt}y^2\right) = 0
$$
or after integration
$$
(y')^2+k^2y^2 = C_1\Rightarrow y' = \pm\sqrt{C_1+k^2y^2}
$$
This DE is separable so
$$
\frac{dy}{\sqrt{C_1+k^2y^2}} = \pm dt
$$
A: $$
                           (\frac{d^2}{dx^2}-k^2)y=0\\
                       (\frac{d}{dx}-k)(\frac{d}{dx}+k)y=0 \\
                            (\frac{d}{dx}+k)y=Ae^{kx} \\
                            e^{-kx}\frac{d}{dx}(e^{kx}y)=Ae^{kx} \\
                              \frac{d}{dx}(e^{kx}y)=Ae^{2kx} \\
                                 e^{kx}y = \frac{A}{2k}e^{2kx}+B \\
                                  y = \frac{A}{2k}e^{kx}+Be^{-kx}.
$$
