I am trying to find analytical solution to the following ODE:
$\dot{x}=c_1x+\frac{c_2}{x}+c_3$, where $c_1,c_2,c_3$ are known constants.
If $c_2=0$, then it's a standard linear ODE which I know how to solve. But, I have no idea how to proceed when all the above constants are non-zero. I really appreciate any hints as to how proceed.
Update: I am looking to get a solution like x=f(t) for the above problem.
Thanks in advance.
$\textbf{P.S.}$: If this information helps, I basically started from the following ODE (below) and applied change of variables to obtain the above form :
$\dot{y}=c_1y+c_3 \sqrt{y}+c_2$ and used $x=\sqrt{y}$ variable transformation to get the above ODE. I want to obtain the analytical solution to this ODE, in fact.
Best Answer
So you have the following differential equation
We can isolate $x^{-1}$ in the right side of the equation to get
$$\frac{\mathrm dx}{\mathrm dt} = x^{-1}(c_1x^2+c_3x+c_2) =: x^{-1}p(x)$$
We can find the roots of the polynom $p(x):= c_1x^2+c_3x+c_2$ using
$$x_{\pm} = \frac{-c_3\pm\sqrt{c_3^2-4c_1c_2}}{2c_1}$$
So we will have that $p(x) = (x-x_{+})(x - x_{-})$ and we get the differential equation, where $x_{+},x_{-} \in \mathbb{R}$ are constants determined by the constants $c_1,c_2,c_3$ given by the problem with the relation above
$$\frac{\mathrm dx}{\mathrm dt} = x^{-1}(x - x_{+})(x - x_{-}) \implies \frac{x}{(x-x_{+})(x - x_{-})} \mathrm dx = \mathrm dt$$
Now we can use that
$$\frac{x}{(x-x_{+})(x - x_{-})} = \frac{A}{x - x_{+}} + \frac{B}{x - x_{-}}$$
Where we have that
$$A := \frac{-x_{+}}{x_{-} - x_{+}} \,\,\,\,\,\,\,\,B:=\frac{x_{-}}{x_{-}-x_{+}}$$
Now we integrate both sides of the equation, if we put $x(0) := x_0$ we will have
$$A\int_{x_0}^x\frac{\mathrm dx'}{x' - x_{+}} + B\int_{x_0}^x\frac{\mathrm dx'}{x' - x_{-}} = A\ln\left(\frac{x - x_{+}}{x_0 - x_{+}}\right) + B\ln\left(\frac{x - x_{-}}{x_0 - x_{-}}\right) = t$$
Then we'll have that
You should work out the case $c_3^2 - 4c_1c_2 < 0$.