Nonlinear differential equation

[PeteSampras]

Homework Statement

I need to solve the follwing differential equation

$$(\frac{df}{dt}) \dfrac{4 n e^{4nf(t)}-9n e^{2nf(t)} (\frac{df}{dt})^2 + e^{2nf(t)} r^2 \frac{d^2f}{dt^2}+5n (\frac{df}{dt})^4 r^4 - r^4 \frac{d^2f}{dt^2} (\frac{df}{dt})^2}{-e^{2nf(t)}+ (\frac{df}{dt})^2 r^2}=0 $$

Homework Equations

r,n are constants >0

The Attempt at a Solution

I tried to solve in Maple the factor

$$4 n e^{4nf(t)}-9n e^{2nf(t)} (\frac{df}{dt})^2 + e^{2nf(t)} r^2 \frac{d^2f}{dt^2}+5n (\frac{df}{dt})^4 r^4 - r^4 \frac{d^2f}{dt^2}(\frac{df}{dt})^2=0$$

but the only solution that i find is such that the denominator is 0.

Also i think in a assumption $\frac{df}{dt} \approx \epsilon$ with $\epsilon^3 \approx 0$ but this is a solution of the form $f(t) = \ln ( g(t) )$ , ¿but the derivative is not small?

Help please

 

[PeteSampras]

f is not constant too

 

[Mark44]

Homework Statement

I need to solve the follwing differential equation

$$(\frac{df}{dt}) \dfrac{4 n e^{4nf(t)}-9n e^{2nf(t)} (\frac{df}{dt})^2 + e^{2nf(t)} r^2 \frac{d^2f}{dt^2}+5n (\frac{df}{dt})^4 r^4 - r^4 \frac{d^2f}{dt^2} (\frac{df}{dt})^2}{-e^{2nf(t)}+ (\frac{df}{dt})^2 r^2}=0 $$

Homework Equations

r,n are constants >0

The Attempt at a Solution

I tried to solve in Maple the factor

$$4 n e^{4nf(t)}-9n e^{2nf(t)} (\frac{df}{dt})^2 + e^{2nf(t)} r^2 \frac{d^2f}{dt^2}+5n (\frac{df}{dt})^4 r^4 - r^4 \frac{d^2f}{dt^2}(\frac{df}{dt})^2=0$$

but the only solution that i find is such that the denominator is 0.

Just to be clear, are you saying that the solution you found also makes the denominator zero?

This is such an ugly differential equation that I don't believe you're going to find an analytic (i.e., exact) solution. Have you tried any numerical methods?

What is the context for this DE?

Also i think in a assumption $\frac{df}{dt} \approx \epsilon$ with $\epsilon^3 \approx 0$

Where does this assumption come from? Is it part of the suggested approach to this problem?

but this is a solution of the form $f(t) = \ln ( g(t) )$ , ¿but the derivative is not small?

Help please

 

[Ray Vickson]

Homework Statement

I need to solve the follwing differential equation

$$(\frac{df}{dt}) \dfrac{4 n e^{4nf(t)}-9n e^{2nf(t)} (\frac{df}{dt})^2 + e^{2nf(t)} r^2 \frac{d^2f}{dt^2}+5n (\frac{df}{dt})^4 r^4 - r^4 \frac{d^2f}{dt^2} (\frac{df}{dt})^2}{-e^{2nf(t)}+ (\frac{df}{dt})^2 r^2}=0 $$

Homework Equations

r,n are constants >0

The Attempt at a Solution

I tried to solve in Maple the factor

$$4 n e^{4nf(t)}-9n e^{2nf(t)} (\frac{df}{dt})^2 + e^{2nf(t)} r^2 \frac{d^2f}{dt^2}+5n (\frac{df}{dt})^4 r^4 - r^4 \frac{d^2f}{dt^2}(\frac{df}{dt})^2=0$$

but the only solution that i find is such that the denominator is 0.

Also i think in a assumption $\frac{df}{dt} \approx \epsilon$ with $\epsilon^3 \approx 0$ but this is a solution of the form $f(t) = \ln ( g(t) )$ , ¿but the derivative is not small?

Help please

As Mark44 has indicated, I think the DE is too nasty to be solved "analytically"; however, in Maple I can get numerical solutions easily. For example, if I set n = 1, r = 0.5 and use initial conditions f(0)=1, f'(0) = 0, the Maple command "sol:=dsolve({de,f(0)=1,D(f)(0)=0},numeric)" has no problem getting a solution that can be plotted out as a function of t. For this solution we can also evaluate the denominator, and it does not equal zero, except for a single root near the right-hand endpoint of the chosen t-interval (0 ≤ t ≤ 3).

In this case I just let Maple choose the default numerical method, but of course you can specify a wide range of methods via options to the dsolve command. Of course, in a serious study one should, in fact, be careful to solve the same system by several competing methods, just to check for accuracy, stiffness effects, stability, etc; and to the extent possible, one should vary the step sizes and maybe the floating-point precision as well, to see how truncation and roundoff errors affect things.

BTW: for the numerical solution in this case, df/dt is not at all small; the solution value f ranges from f(0) = 1 to about f(3) = -0.8, and the derivative is not quite linear over the range t = 0 --> 3; that is, df/dt varies with an average value of about -0.6.