Any approximate analytical solution to this ode?

[albertshx]

I'm working on this differential equation this few days... Could you give
some guidance on approximate solutions to it? i(t) is the only function
while all others are parameters.

$$\frac{di(t)}{dt} = -\lambda(\sigma\phi\sqrt{i(t)(1-i(t))} + N\mu i(t)(1-i(t))$$

Thank you a lot!

 

[JJacquelin]

The equation is not clearly written.
It seems that this ODE is on the "separable" kind. You may apply the classical method of resolution for this kind of equations.

 

[3.1415926535]

\begin{array}{l}\frac{di(t)}{dt} = -\lambda(\sigma\phi\sqrt{i(t)(1-i(t))} + N\mu i(t)(1-i(t))\\
\frac{di(t)}{dt} = -\lambda\sigma\phi\sqrt{i(t)(1-i(t))} -N\lambda\mu i(t)(1-i(t))\\
\
-\lambda\sigma\phi=u\\
-N\lambda\mu=v\\
\frac{di(t)}{dt} = u\sqrt{i(t)(1-i(t))}+ vi(t)(1-i(t))\\
\frac{di(t)}{u\sqrt{i(t)(1-i(t))}+ vi(t)(1-i(t))} =dt \\
\int\frac{di(t)}{u\sqrt{i(t)(1-i(t))}+ vi(t)(1-i(t))}=\int dt\\
\int\frac{di(t)}{u\sqrt{i(t)(1-i(t))}+ vi(t)(1-i(t))}=t+c\\
t=\int\frac{di(t)}{u\sqrt{i(t)(1-i(t))}+ vi(t)(1-i(t))}-c\end{array}

Now you have to calculate that integral...
The solution from Wolfman Mathematica is the attached image(Log[x] is the natural log of x and tan^(-1)(x)=arctanx is the inverse trigonometric tan of x)

Replace x with i(t) and you have the inverse function, t(i). Finding the i(t) is up to you.

 

[albertshx]

Many, many thanks! You are much more careful than me. I just tried with MATLAB and dsolve reports explicit solution not found. So I look for some approximation.