2nd order differentiel equation

Question

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 ?

Answer 1

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.