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Solve the $$x'=tx^2+x-t^3\,,\quad x\left(\, 2\,\right)=1$$

I need its exact solution not a numerical solution.In fact I have to compare the exact solution with the numerical solution.I tried it but I am unable to solve it for the previous one day.I need your kind help.

Thanks.

Felix Marin
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user163993
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  • This is a Riccati ODE. Pretty sure you're not going to find a solution in terms of elementary functions for your particular IVP. – Simon S Dec 19 '14 at 17:03
  • But unfortunately I need the exact solution. – user163993 Dec 19 '14 at 17:06
  • can we compute $x(2.2)$ from here? – user163993 Dec 19 '14 at 17:07
  • Then you might it helpful to start here: http://en.wikipedia.org/wiki/Riccati_equation – Simon S Dec 19 '14 at 17:08
  • I have computed $x(2.2)$ with Runge-Kutta method of order 4 and the fourth order taylor series method.I have to compare these with exact value of $x(2.2)$ – user163993 Dec 19 '14 at 17:09
  • No I have already calculated $x(2.2)$ with Runge-Kutta method of order 4 and the fourth order taylor series method and I have to compare both these with the exact value of $x(2.2)$ – user163993 Dec 19 '14 at 17:12
  • Typically with Riccati's you can try to find ONE solution to it and then build the general solution from there(good luck!). Otherwise, you might as well use a numerical technique stronger than RK4 to get a more precise answer. It may not be the intention of the exercise to find an analytic solution. – DaveNine Aug 30 '15 at 18:15

1 Answers1

2

Hint:

Let $x=-\dfrac{u'}{tu}$ ,

Then $x'=-\dfrac{u''}{tu}+\dfrac{u'}{t^2u}+\dfrac{(u')^2}{tu^2}$

$\therefore-\dfrac{u''}{tu}+\dfrac{u'}{t^2u}+\dfrac{(u')^2}{tu^2}=\dfrac{(u')^2}{tu^2}-\dfrac{u'}{tu}-t^3$

$\dfrac{u''}{tu}-\dfrac{u'}{tu}-\dfrac{u'}{t^2u}-t^3=0$

$tu''-(t+1)u'-t^5u=0$

Let $u=e^{at^3}v$ ,

Then $u'=e^{at^3}v'+3at^2e^{at^3}v$

$u''=e^{at^3}v''+3at^2e^{at^3}v'+3at^2e^{at^3}v'+(9a^2t^4+6at)e^{at^3}v=e^{at^3}v''+6at^2e^{at^3}v'+(9a^2t^4+6at)e^{at^3}v$

$\therefore t(e^{at^3}v''+6at^2e^{at^3}v'+(9a^2t^4+6at)e^{at^3}v)-(t+1)(e^{at^3}v'+3at^2e^{at^3}v)-t^5e^{at^3}v=0$

$t(v''+6at^2v'+(9a^2t^4+6at)v)-(t+1)(v'+3at^2v)-t^5v=0$

$tv''+(6at^3-t-1)v'+((9a^2-1)t^5-3at^2(t-1))v=0$

Choose $a=\dfrac{1}{3}$ , the ODE becomes

$tv''+(2t^3-t-1)v'-t^2(t-1)v=0$

Let $v=e^{bt^2}w$ ,

Then $v'=e^{bt^2}w'+2bte^{bt^2}w$

$v''=e^{bt^2}w''+2bte^{bt^2}w'+2bte^{bt^2}w'+(4b^2t^2+2b)e^{bt^2}w=e^{bt^2}w''+4bte^{bt^2}w'+(4b^2t^2+2b)e^{bt^2}w$

$\therefore t(e^{bt^2}w''+4bte^{bt^2}w'+(4b^2t^2+2b)e^{bt^2}w)+(2t^3-t-1)(e^{bt^2}w'+2bte^{bt^2}w)-t^2(t-1)e^{bt^2}w=0$

$t(w''+4btw'+(4b^2t^2+2b)w)+(2t^3-t-1)(w'+2btw)-t^2(t-1)w=0$

$tw''+(2t^3+4bt^2-t-1)w'+(4bt^4+(4b^2-1)t^3-(2b-1)t^2)w=0$

Choose $b=\dfrac{1}{2}$ , the ODE becomes

$tw''+(2t^3+2t^2-t-1)w'+2t^4w=0$

doraemonpaul
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