91Ó°ÊÓ

The differential equation is $y^{\text{'}\text{'}2}=k^2\left(1+y^{\text{'}2}\right)$.

A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of a function at a given moment define its rate of change.

The given equation is lacking by independent variable, thus it can be significantly simplified by using the substitution, $y^{\text{'}}≡p$also implying by $y^{\text{'}\text{'}}=p^{\text{'}}$exploiting the chain rule for derivatives. Notice that the equation lacking the variable x as well, but that this transformation leads to significantly simpler solution.

Apply the square root operation on the equation
$y^{\text{'}\text{'}2}=k^2\left(1+y^{\text{'}2}\right)⇒y^{\text{'}\text{'}}=Â\pm k\sqrt{1+y^2}$
Notice the sign since the signature of the constant is not given.
$p^{\text{'}}=Â\pm k\sqrt{1+p^2}$
And since the previous equation is manifestly separable, so by separating it to obtain:
$$\begin{gathered} \frac{dp}{dx}=Â\pm k\sqrt{1+p^2} \\ ⇒\frac{dp}{\sqrt{1+p^2}}=Â\pm kdx \end{gathered}$$
Bearing in mind ${\cosh }^{2}x={\sinh }^{2}x+1$, and substitute the $p=\sinh u⇒dp=\cosh udu$
$$\begin{gathered} \frac{dp}{\sqrt{1+p^2}}=Â\pm kdx \\ \frac{\cosh udu}{\sqrt{1+\sinh u^2}}=Â\pm kdx \\ \frac{\cosh udu}{\cosh u}=Â\pm kdx \end{gathered}$$

By integration and using formula $∫dx=x+C$then the previous equation

$$\begin{gathered} u=Â\pm kx+A \\ ⇒{\sinh }^{-1}p=Â\pm kx+A \\ ⇒p=\sinh (Â\pm kx+A) \end{gathered}$$
Since$p=\frac{dy}{dx}$ so they can integrate once again and use the table integral
$$\begin{gathered} ∫\sinh (ax+b)dx=\frac{1}{a}\cosh (ax+b)+C \\ y=\frac{1}{Â\pm k}\cosh (Â\pm kx+A)+B \end{gathered}$$
Thus, the solutions of the differential equation$y=\frac{1}{Â\pm k}\cosh (Â\pm kx+A)+B$