Solving a differential equation

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SUMMARY

The discussion centers on solving the differential equation given by Equation (1): \(\frac{dx}{dt}=1+x^2w^2\). The proposed solution is Equation (2): \(x(t)=\frac{Ax(0)+B}{Cx(0)+D}\), which raises questions about its validity since \(x(0)\) is a constant. A participant suggests using a trigonometric substitution to integrate Equation (3): \(\frac{dx}{1+x^2w^2}=dt\), leading to the conclusion that Equation (2) holds true if the constants A, B, C, and D are functions of time.

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  • Knowledge of initial conditions in the context of differential equations.
  • Basic concepts of functions and their dependence on variables.
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Mathematicians, physics students, and anyone involved in solving differential equations, particularly those interested in integration techniques and the analysis of solutions with variable parameters.

zinDo
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I want to solve Equation (1). w is a constant:
\begin{eqnarray} \text{Equation (1): }\frac{dx}{dt}=1+x^2w^2\end{eqnarray}
and I have been told that it is solved by (2):
\begin{eqnarray} \text{Equation (2): }x(t)=\frac{Ax(0)+B}{Cx(0)+D}\end{eqnarray}

Problem

I believe them, but before I keep solving it I want to know how one concludes that (1) is solved by (2). In order to get (2), should I integrate something like (3)? Is it some kind of ansatz that I should have known? ...
\begin{eqnarray} \text{Equation (3): }\frac{dx}{1+x^2w^2}=dt\end{eqnarray}
I don't know what is it that I should be looking for. Could I get a clue?
Thank you very much
 
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zinDo said:
I want to solve Equation (1). w is a constant:
\begin{eqnarray} \text{Equation (1): }\frac{dx}{dt}=1+x^2w^2\end{eqnarray}
and I have been told that it is solved by (2):
\begin{eqnarray} \text{Equation (2): }x(t)=\frac{Ax(0)+B}{Cx(0)+D}\end{eqnarray}
This doesn't make sense to me, assuming that I'm interpreting your notation correctly. x(0) is just a constant, so what you have on the right side in equation 2 is just a constant (doesn't depend on t). It couldn't possibly be a solution to your diff. equation. x(t) = K, so dx/dt = 0, and 1 + x2w2 ≥ 1.
zinDo said:
Problem

I believe them, but before I keep solving it I want to know how one concludes that (1) is solved by (2). In order to get (2), should I integrate something like (3)? Is it some kind of ansatz that I should have known? ...
\begin{eqnarray} \text{Equation (3): }\frac{dx}{1+x^2w^2}=dt\end{eqnarray}
I don't know what is it that I should be looking for. Could I get a clue?
I would use a trig substitution to integrate what you have on the left side just above.
 
##\int (\frac{1}{1+x^2}dx)=tan^{-1}(x) +C## , one of the basic integrals.

Eq. (2) is true if A, B, C, D are functions of t.
 

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