I have the following ODE: ($y:[0,1]\rightarrow \mathbb{R} , t\mapsto y(t)$ a function) $$y''(t) -t y'(t) -2 y(t) = 0, y(0)=1, y'(0)=a\in \mathbb{R}.$$

Because there is a factor $t$ in front of $y'(t)$ this equation does not have constant coefficients. I found a recipe here: http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf of how to solve 2nd order ODE's but this case seems to be difficult.

I can't find any particular solution to this equation. Does anyone know how to solve it or proceed? Any hint is highly appreciated.


  • $\begingroup$ What have you tried? For example, have you put it into the formally symmetric Sturm-Liouville form $-(py')'+qy=0$? Can you perform a variation of parameters calculation, treating $qy$ as the inhomogeneity? $\endgroup$
    – Stromael
    Mar 9, 2015 at 12:55
  • $\begingroup$ According to the notes I was trying to find a particular solution, of the form $at+b$ but it didn't work. $\endgroup$
    – Martingalo
    Mar 9, 2015 at 13:02
  • $\begingroup$ Apparently, $y(t)= C te^{t^2/2}$ solves the equation with $C\in \mathbb{R}$ but does not satisfy the boundary conditions. $\endgroup$
    – Martingalo
    Mar 9, 2015 at 13:24
  • $\begingroup$ Given one solution, there is a general method to find another independent solution... $\endgroup$
    – GEdgar
    Mar 9, 2015 at 13:46
  • $\begingroup$ That method is given in the last couple of pages, of the linked material. $\endgroup$
    – mvw
    Mar 9, 2015 at 13:49

1 Answer 1


The general solution for that DE is $$ A \;t \exp\left(\frac{t^2}{2}\right) + B\left(\sqrt{2\pi}+t\exp\left(\frac{t^2}{2}\right)\;\pi\;\mathrm{erf}\left(\frac{t}{\sqrt{2}}\right)\right) $$ Use your initial conditions to find the constants $A,B$.

  • $\begingroup$ Oh! How did you get to that? :O $\endgroup$
    – Martingalo
    Mar 10, 2015 at 17:41
  • $\begingroup$ See mvw's comment to the original question. $\endgroup$
    – GEdgar
    Mar 10, 2015 at 17:44

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