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I was trying to find out the solutions of the following differential equation:

\begin{equation}2 \frac{dA}{dt} (\sqrt{B^{2}-D})-2\frac{dB}{dt}(B)+\frac{dD}{dt}=0\end{equation}

where $A,B,D \in \textit{C}^{\infty}$.

Suggestions?

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    Let me make sure I understand the question: You have _one_ differential equation, with three dependent variables (A,B,D) in it, and you want to solve for the three of them from that one ODE? – 2017-02-20
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    if it's possible, yes. There might be classes of infinite solutions, which is fine anyway. I'm wondering if there is some way to find them analytically (in this case I mean) – 2017-02-20
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    ... and you are sure, that you want to ask this here instead of Math.SE? – 2017-02-20
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    It can be easily seen that any functions `a[t], b[t], c[t]` such that `a[t]-Sqrt[b[t]^2-d[t]] == const` satisfy this equation identically. Here `const` means that it doesn't depend on `t`. – 2017-02-20

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Well, think about the three-space $\mathbb{R}^3$ with coordinates $p = (A,B,D)\, \in \, \mathbb{R}^3$. Then, if you ignore the parametrization $t \in \mathbb{R}$ for a moment, you realize you have a differential one-form \begin{equation} \Omega_{p} = 2\, (\sqrt{B^{2}-D}) \, dA - 2\,B \, dB + dD = 0\end{equation} At each point $p = (A,B,D)\, \in \, \mathbb{R}^3$ the equation $\Omega_p = 0$ defines a two-dimensional tangent plane, which when $p$ moves around the three space gives you a field of tangent planes $\{F_p \, : \, p \in \mathbb{R}^3\}$, also called a two-dimensional distribution. The question is, is there a curve $p = p(t) = (A(t), B(t), D(t))$ which is tangent to the plane field $F_p$.

In general, the first thing you could check is whether the distribution $F_p$ is integrable (meaning whether it defines a foliation of two dimensional surfaces tangent to the plane field) by checking the conditions of Frobenius' theorem, which in this case means to check whether $ \, \Omega \, \wedge \, d\Omega = 0$. In your case however, the distribution is fairly simple and explicitly integrable. Indeed

$$0 = 2\, (\sqrt{B^{2}-D}) \, dA - 2\,B \, dB + dD = 2\, (\sqrt{B^{2}-D}) \, dA - d(B^2) + dD = $$ $$= 2\, (\sqrt{B^{2}-D}) \, dA - d(B^2 - D) $$ Now divide both sides by $ 2\, (\sqrt{B^{2}-D})$ obtaining the equivalent equation $$dA \, -\, \frac{d(B^2-D)}{2\, (\sqrt{B^{2}-D})} = 0$$ $$dA \, -\, d\Big(\, \sqrt{(B^2-D)}\, \Big) = 0$$ leading to the solution $$A = \sqrt{(B^2-D)}\, + \, c_0$$ where $c_0$ is a real constant. So if you pick any smooth functions $B=B(t), \, D = D(t)$, that do not violate the restrictions $B(t)^2 \geq D(t)$ of the square root, you get $$A(t) = \sqrt{\Big(B^2(t)-D(t)\Big)}\, + \, c_0$$ with an arbitrary constant $c_0$.