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Solving a differential equation

  • #1
2
0
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
 

Answers and Replies

  • #2
33,285
4,989
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.
 
  • #3
ehild
Homework Helper
15,427
1,827
##\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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