laura123 said:
yes $\chi\sigma$, it seems simple...apparently
May be it seems... but in fact it is hard!... the second order ODE...
$\displaystyle y^{\ ''} + y\ \frac{x^{3}}{1+x^{3}}\ (1)$
... is linear homogeneous and that means that the solution is of the form...
$\displaystyle y(x) = c_{1}\ u(x) + c_{2}\ v(x)\ (2)$
... where u(*) and v(*) are independent solutions of (1). An asyntotic solution for x>>1 of course is...
$\displaystyle y(x) \sim c_{1}\ \cos x + c_{2}\ \sin x\ (3)$
... but what what does it happen for more small values of x?... one attempt to approximate the solution is to suppose that y(x) is analytic in x=0, so that is...
$\displaystyle y(x)= a_{0} + a_{1}\ x + a_{2}\ x^{2} + ...\ (4)$
For semplicity we suppose also that is $y(0)= 1 \implies a_{0}=1$ and $y^{\ '}(0)=0 \implies a_{1}=0$ so that the problem is to compute the succesive derivatives. From (1) we derive directly...
$\displaystyle y^{\ ''} = - y\ \frac{x^{3}}{1+x^{3}} = 0\ \text{in}\ x=0\ (5)$
Proceeding from (5) we have...
$\displaystyle y^{(3)} = - 3\ y\ \frac{x^{2}}{(1+x^{3})^{2}} - y^{\ '}\ \frac{x^{3}}{1+x^{3}}= 0\ \text{in}\ x=0\ (6) $
$\displaystyle y^{(4)} = y\ \frac{12\ x^{2} - 6\ x}{(1+x^{3})^{3}} - y^{\ '}\ \frac{6\ x^{2}}{(1+x^{3})^{2}} - y^{\ ''}\ \frac{x^{3}}{1+x^{3}} = 0\ \text{in}\ x=0\ (7) $
$\displaystyle y^{(5)} = - 6\ y\ \frac{14\ x^4 - 8\ x^{3} - 4\ x + 1}{(1 + x^{3})^{4}} + ... = -6\ \text{in}\ x=0\ (7)$
If we stop now the approximate solution near x=0 is...
$\displaystyle y(x) \sim 1 - \frac{x^{5}}{20}\ (8)$
If more precision is requested we can go ahead but the task is very inconfortable... at least for me (Worried)... A numerical solution is clearly preferable...
Kind regards
$\chi$ $\sigma$
