Solve the given initial value problem.I need your help. 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.
 A: 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
$tw''+(2t^3+2t^2-t-1)w'+2t^4w=0$
