Solving $y''-k^2y=0$ without substituting $e^{kx}$

Question

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.

Answer 1

Hint: try hyperbolic substitution $$y=a\sinh u$$or$$y=a\cosh u$$ for some proper constant $a$

Answer 2

$$\frac{dy}{dx}=\sqrt{k^2y^2+c}$$ 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 $$y(x)=c_1e^{kx}+c_2e^{-kx}$$

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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 $z=\frac {y'}y$

Answer 3

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

$$ \frac{dy}{\sqrt{C_1+k^2y^2}} = \pm dt $$

Answer 4

$$ (\frac{d^2}{dx^2}-k^2)y=0\\ (\frac{d}{dx}-k)(\frac{d}{dx}+k)y=0 \\ (\frac{d}{dx}+k)y=Ae^{kx} \\ e^{-kx}\frac{d}{dx}(e^{kx}y)=Ae^{kx} \\ \frac{d}{dx}(e^{kx}y)=Ae^{2kx} \\ e^{kx}y = \frac{A}{2k}e^{2kx}+B \\ y = \frac{A}{2k}e^{kx}+Be^{-kx}. $$