15
$\begingroup$

I need to solve the differential equation $$\frac{dy(t)}{dt}=C_0+C_1y+C_2y^2+C_3y^3$$ where $C_i$ are constants.

My attempt:

By separation of variables I have $$\int_{a}^{y}\frac{dy}{\frac{C_0}{C_3}+\frac{C_1}{C_3}y+\frac{C_2}{C_3}y^2+y^3}=C_3\int_{0}^{t}dt$$ If $r_1$, $r_2$, and $r_3$ denote the three roots of the polynomial in the denominator of the above relation, I obtain $$\int_{a}^{y}\frac{A_1dy}{y-r_1}+\int_{a}^{y}\frac{A_2dy}{y-r_2}+\int_{a}^{y}\frac{A_3dy}{y-r_3}=C_3t$$ from which it results $$A_1\ln(\frac{y-r_1}{a-r_1})+A_2\ln(\frac{y-r_2}{a-r_2})+A_3\ln(\frac{y-r_3}{a-r_3})=C_3t$$

Using properties of $\ln$ function, this simplifies to $$\ln\left[(\frac{y-r_1}{a-r_1})^{A_1}(\frac{y-r_2}{a-r_2})^{A_2}(\frac{y-r_3}{a-r_3})^{A_3}\right]=C_3t$$ but I can't extract $y$ as an explicit function of $t$ from the above equation.

Does anyone have any idea about this ODE and the approach that must be taken to solve it? If it's not solvable analytically, could anyone suggest some good methods to approximate the analytical (not numerical) solution of this ODE, please?

(There isn't any restriction on the kind of the solution functions, i.e. the solution could be in terms of special functions)

$\endgroup$
1

2 Answers 2

8
$\begingroup$

There is no closed form (explicit) solution of this ODE, you cannot express $y$ as a function of $t$ in terms of common functions. Your approach although to finding an expression for the solution is indeed correct and that's probably one of the most simplified expressions you can yield for the solution $y(t)$ of the problem. Good job!

Note that such problems can be solved using numerical methods.

As for algebraic ways, as you've figured, there isn't always a way to find closed form solutions, that's exactly why we have Numerical Methods. But, take into account that an expression involving $y(t)$ without any derivative forms of it is considered a solution of sort and is totally acceptable in such cases.

$\endgroup$
6
$\begingroup$

$$\frac{dy(t)}{dt}=C_0+C_1y+C_2y^2+C_3y^3$$ You have correctly solved the ODE. $$A_1\ln|\frac{y-r_1}{a-r_1}|+A_2\ln|\frac{y-r_2}{a-r_2}|+A_3\ln|\frac{y-r_3}{a-r_3}|=C_3t$$ The solution is expressed on the form of an implicit equation.

The ODE is solved whatever the form given for the solution : parametric, or implicit, or explicit.

If the wording of the problem specifies that the solution must be explicit, the further problem is of different kind : This is an invertion problem.

Let $\quad f(y)=\frac{A_1}{C_3}\ln|\frac{y-r_1}{a-r_1}|+\frac{A_2}{C_3}\ln|\frac{y-r_2}{a-r_2}|+\frac{A_3}{C_3}\ln|\frac{y-r_3}{a-r_3}|$

Formally the solution is denoted $$y=f^{-1}(t)$$ This is the explicit solution on formal sense. But of course this is not satisfising on the common sense of "explicit".

Note that asking for "explicit" form supposes to specify what kind of functions are allowed to express the result : elementary functions, standard special functions, unstandardized special functions, etc. and to specify if infinite series are or not allowed. This doesn't appear in the present wording of the question.

$\endgroup$
2
  • $\begingroup$ There isn't any restriction on the kind of functions that could represent the answer. You mean that there are some kind of special functions that $y(t)$ might be expressed by them? Could you please specify what are these special functions? $\endgroup$
    – Masa
    Nov 3, 2019 at 10:53
  • $\begingroup$ If there is no restriction at all, one can define a non-standard special function as $$\quad f(x)=\frac{A_1}{C_3}\ln|\frac{x-r_1}{a-r_1}|+\frac{A_2}{C_3}\ln|\frac{x-r_2}{a-r_2}|+\frac{A_3}{C_3}\ln|\frac{x-r_3}{a-r_3}|$$ and the analytical solution is $$y=f^{-1}(t)$$. Of course this appears just as a trick without interest. But that is implied by "without any restriction". In fact everybody understand that they are some implicit restrictions. But with them your question about explicit solution has no answer. As already said, many ODEs are solvable but not explicitly.. $\endgroup$
    – JJacquelin
    Nov 3, 2019 at 11:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .