Given $$\frac{d^2u}{dt^2}-4t\frac{du}{dt}+(4t^2-2)u=0$$ Find the general solution.

I believe I'm supposed to find the fundamental solutions first, by rewriting the equation into two $1^{st}$ order differential equations, and I have done so. However, this will lead to pretty ugly eigenvalues ($\lambda_{1,2}=2t\pm\sqrt{8t^2-2}$ and therefore ugly eigenvectors. Am I supposed to do it this way or is there an easier/quicker method?

Update Using Wolframalpha I've managed to find a solution: $$y(t)=c_1e^{t^2}+c_2te^{t^2}$$ However, I would like to know why this is so.


1 Answer 1


Apply the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf:

Let $u=e^{t^2}v$ ,

Then $\dfrac{du}{dt}=e^{t^2}\dfrac{dv}{dt}+2te^{t^2}v$


$\therefore e^{t^2}\dfrac{d^2v}{dt^2}+4te^{t^2}\dfrac{dv}{dt}+(4t^2+2)e^{t^2}v-4t\left(e^{t^2}\dfrac{dv}{dt}+2te^{t^2}v\right)+(4t^2-2)e^{t^2}v=0$






  • 1
    $\begingroup$ Is the use of $u=e^{t^2}v$ an educated guess or is there some sort of theory that describes why this is 'the way to go'? $\endgroup$
    – Di-lemma
    Apr 2, 2016 at 18:17

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