Linear equation solving

Question

I cant solve $tx'+\dfrac{tx}{\sqrt{1+t^3}}=1$ I have tried to do it like an homogenian but i cant integrate $\dfrac{1}{\sqrt{1+t^3}}$ so i suposse it must be done by another method

Answer 1

I wasn't sure, from reading the OP's question, exactly where he got stuck, although certainly the integrals to which this question leads are difficult

This being the case:

The following shows how the solution to $tx' + \frac{tx}{\sqrt{1 + t^3}} = 0$ may be expressed in terms of integrals of the known functions $(\sqrt{1 + t^3})^{-1}$, $t^{-1}$, which I don't know how to evaluate in closed form. But it does show the general procedure for reducing equations of the form $x' + p(t)x = q(t)$ to quadratures.

Let's assume $t \ne 0$, since the case $t = 0$ makes no sense, reverting as it does to the contradictory assertion that $0 = 1$. Then dividing through by $t \ne 0$ we obtain

$x' + \frac{x}{\sqrt{1 + t^3}} = t^{-1}. \tag{1}$

This equation is of the general form

$x' + p(t)x = q(t) \tag{2}$

with

$p(t) = \frac{1}{\sqrt{1 + t^3}} \tag{3}$

and

$q(t) = t^{-1}; \tag{4}$

as such, there are explicit and relatively simple formulas for the solution $x(t)$, though they may contain integrals which are difficult to evaluate. Nevertheless, they allow a solution to be expressed in relatively simple form, "reduced to quadratures", i.e. in terms of integrals over $t$ involving the presumably known functions $p(t), q(t)$. To apply them, initial conditions for (1) are necessary, so let us assume $x = x_0$ at $t = t_0$.

We first address the homogeneous variant of (1), viz.

$x' + p(t)x = 0, \tag{5}$

which can easily be transformed to

$\frac{x'}{x} = - p(t), \tag{6}$

as long as $x \ne 0$. This is a safe assumption in this case, since the linearity of (5) implies that the only solution passing through $0$ is in fact $x(t) = 0$. (This assertion in fact requires uniqueness of solutions for (5), but this follows readily from the fact that (5), being linear, is easily seen to satisfy the requisite Lipschitz continuity.) From (6) we have

$(\ln x)' = -p(t), \tag{7}$

and if we integrate using $x = x_0$ at $t = t_0$ we have

$\ln x(t) - \ln x_0 = -\int_{t_0}^t p(s)ds, \tag{8}$

or

$\ln (\frac{x(t)}{x_0}) = -\int_{t_0}^t p(s)ds, \tag{9}$

or, finally,

$x(t) = x_0 e^{ -\int_{t_0}^t p(s)ds}. \tag{10}$

The exponential occurring in (10) can immediately be used to develop a solution $x(t)$ for (2) such that $x_0 = 0$; indeed, we have, upon multiplying (2) by $e^{ +\int_{t_0}^t p(s)ds} = (e^{-\int_{t_0}^t p(s)ds})^{-1}$,

$(e^{\int_{t_0}^t p(s)ds}x(t))' = e^{\int_{t_0}^t p(s)ds}(x' + p(t)x) = e^{\int_{t_0}^t p(s)ds}q(t) \tag{11}$

and integrating (11) yields

$e^{\int_{t_0}^t p(s)ds}x(t) - x_0 = e^{\int_{t_0}^t p(s)ds}x(t) = \int_{t_0}^t e^{\int_{t_0}^r p(s)ds}q(r)dr, \tag{12}$

since we're taking $x_0 = 0$ for the purposes of the present calculation. (12) yields

$x(t) = e^{-\int_{t_0}^t p(s)ds} \int_{t_0}^t e^{\int_{t_0}^r p(s)ds}q(r)dr, \tag{12}$

the general solution of (2) with $x_0 = 0$. If we add to this our solution to the homogeneous equation (5) we obtain

$x(t) = x(t) = x_0 e^{ -\int_{t_0}^t p(s)ds} + e^{-\int_{t_0}^t p(s)ds} \int_{t_0}^t e^{\int_{t_0}^r p(s)ds}q(r)dr, \tag{13}$

which solves (2) with $x = x_0$ at $t = t_0$.

So much for the general formula. Inserting (3) and (4) into (13):

$x(t) = x(t) = x_0 e^{ -\int_{t_0}^t \frac{1}{\sqrt{1 + s^3}}ds} + e^{-\int_{t_0}^t \frac{1}{\sqrt{1 + s^3}}ds} \int_{t_0}^t e^{\int_{t_0}^r \frac{1}{\sqrt{1 + s^3}}ds}r^{-1}dr, \tag{14}$

which reduces (1) to quadratures. In principle, we have a solution at this point, though the integrals occurring in (14) seem to be difficult to express in closed form, if indeed it is possible at all. However, having (14) at our disposal allows the use of numerical integration to obtain accurate approximations to the true solution. Another caveat which should be mentioned is that the interval of integration, $[t_0, t]$, must be chosen so as not to contain either $-1$ or $0$, where $p(t)$, $q(t)$ respectively are singular, i.e., are undefined.

That's about as far as I can take it at the present time.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!