<s>[INST] 2nd order differentiel equation I was wondering how I could solve
$$
f''=\left(1+x^4\right)f
$$
for some $f$.
I must admit that when it is 2nd order depending on $x$ I've more struggle to succeed, but here i've no clue ! Any hint please ?
 [/INST] Hint:
Let $f=e^{ax^3}u$ ,
Then $f'=e^{ax^3}u'+3ax^2e^{ax^3}u$
$f''=e^{ax^3}u''+3ax^2e^{ax^3}u'+3ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u=e^{ax^3}u''+6ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u$
$\therefore e^{ax^3}u''+6ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u=(1+x^4)e^{ax^3}u$
$u''+6ax^2u'+(9a^2x^4+6ax)u=(x^4+1)u$
$u''+6ax^2u'+((9a^2-1)x^4+6ax-1)u=0$
Choose $9a^2-1=0$ , i.e. $n=\dfrac{1}{3}$ , the ODE becomes
$u''+2x^2u'+(2x-1)u=0$
Let $t=bx$ ,
Then $b^2\dfrac{d^2u}{dt^2}+\dfrac{2t^2}{b}\dfrac{du}{dt}+\left(\dfrac{2t}{b}-1\right)u=0$
$\dfrac{d^2u}{dt^2}+\dfrac{2t^2}{b^3}\dfrac{du}{dt}+\left(\dfrac{2t}{b^3}-\dfrac{1}{b^2}\right)u=0$
Choose $b^3=2$ , i.e. $b=\sqrt[3]2$ , the ODE becomes
$\dfrac{d^2u}{dt^2}+t^2\dfrac{du}{dt}+\left(t-\dfrac{1}{\sqrt[3]4}\right)u=0$
Which relates to Heun's Triconfluent Equation.
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