I am asked to use the variation of parameters method to find the particular solution :

$$t^2y'' - t(t+2)y' + (t+2)y = 6t^3 \\ t>0 \\ y_1(t) =t \\ y_2(t) = te^t$$

Where $y_i$ are the solutions of the homogenous one. I am using method of variation of parameters so I have the following formulas in my mind :

$$u_1' = \frac{-y_2g}{W(y_1,y_2)} \ u_2' = \frac{y_1 g}{W(y_1,y_2)}$$

where $W(y_1,y_2)$ is the wronskian.

Then clearly $u_1 = -6t + c_1 \ u_2 = -6e^{-t} + c_2$ So the general solution is :

$$y= (-6t+c_1)(t) + (-6te^{-t}+c_2)(te^t) \\ = tc_1+te^{t}c_2-6t^2-6t$$

So I can take $-6t^2-6t$ as the particular solution. However answer sheet says that the answer is $-6t^2$ I kept checking my calculation mistakes on and on and on but couldn't find any? What am I doing wrong?

  • 1
    $\begingroup$ Let $k_1:=c_1-6$. Note that $k_1$ is an arbitrary constant, just like $c_1$ is. $\endgroup$ – projectilemotion Jul 17 '17 at 18:37

Your particular solution is in fact correct. Any two particular solutions agree if their difference solves the homogeneous equation. Here the difference is $6t$ which is a constant times the quoted solution for $y_1$.


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