Can a Riccati-type Differential Equation with a Square Root Term be Classified?

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SUMMARY

The discussion centers on a second-order differential equation (DE) of the form Q''[t] = f[t]Q'[t] + g[t]Q[t] + h[t]*Sqrt[(Q[t])^2 + (k[t]Q'[t])^2]. The presence of the square root term complicates its classification, as it deviates from standard Riccati-type equations. The transformation Q(t)=e^{\int_0^t dx G(x)} simplifies the equation to a first-order nonlinear DE, G'(t)=-G(t)^2+f(t) G(t)+g(t)+h(t) Sqrt[G(t)^2 k(t)^2+1]. Further exploration of the functions f(t), g(t), h(t), and k(t) is essential for potential solutions.

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Mathematicians, researchers in applied mathematics, and students studying differential equations who are interested in advanced classification and solution techniques for complex DEs.

Elros
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Hi Folks..I came up with a weird looking DE for my research

Q''[t] = f[t]Q'[t] + g[t]Q[t] + h[t]*Sqrt[(Q[t])^2 + (k[t]Q'[t])^2]

The thing that mixes my mind is the square root term..If there is no square root terms it would like a riccati type..but there is..i tried to google it but it just showed me bunch of papers about fractional power DEs which is different...Is there any classification for this type ?

Thanks..
 
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Common any ideas ?
 
Actually, setting:

Q(t)=e^{\int_0^t dx G(x)}

reduces your 2nd order ODE to a 1st order nonlinear one:

G'(t)=-G(t)^2+f(t) G(t)+g(t)+h(t) \sqrt{G(t)^2 k(t)^2+1}

Depending on the form of the functions f(t),h(t), g(t) and k(t) you might be able to solve the latter and then again it might be worse...
 

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