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Good evening everyone ,

Could you please explain to me how we solve the following linear differential equation with constant coefficients : $ (E) \ \ y^{(n)} - y = e^{ \alpha x} $ with $ \alpha $ an $n$- root of unity ? The solutions of this equation is put in the form: $ y = y_G + y_P $ with : $ y_G $ general solution of the equation $ y^{(n)} - y = 0 $ is of the form: $ y_{G} (x) = \displaystyle \sum_{ i = 0}^{ n-1 } \lambda_{i} e^{ \mu^{i} x} $ with $ \mu $ is an $ n $ th root of unity and $ \lambda_i \in \mathbb{R} $ for all $ i = 0, 1, \dots , n-1 $. In contrast, I do not know how to find $ y_{P} $ : particular solution of the equation $ (E) $.

Thank you in advance for your help.

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$\bf hint:$ make a change of variable $y = ve^{\alpha x}.$ find and solve the differential equations satisfied by $v.$

$\bf edit:$

you only need the coefficient of $v'.$ just to be specific, let us take $n = 3.$ then $$y' = (v' + \alpha v)e^{\alpha x}, y'' = (v'' + 2 \alpha v' + \alpha^2 v)e^{\alpha x}, y''' = (v''' + 3 \alpha v'' + 3 \alpha^2 v' + \alpha^3 v)e^{\alpha x}$$ we need to solve $$e^{\alpha x} = y''' - y = e^{\alpha x}\left(v''' + 3 \alpha v'' + 3 \alpha^2 v' + \alpha^3 v - v \right)$$ that is $$v''' + 3 \alpha v'' + 3 \alpha^2 v' = 1$$ this has a particular solution $$v = \frac 1{3\alpha^2}x, $$

$$ y = \frac 1{3\alpha^2}xe^{\alpha x} \text{ is a particular solution of } y''' - y = e^{\alpha x} \text{ where $\alpha$ is a cube root of $1.$ } $$

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  • $\begingroup$ $ v=v(x) $, right ? $\endgroup$ – Bryan261 Mar 14 '15 at 11:41
  • $\begingroup$ @Bryan261,yes. $v$ is a function of $x.$ $\endgroup$ – abel Mar 14 '15 at 11:42
  • $\begingroup$ Thank you. This is what i obtain finally : if : $ y(x) = v(x) e^{ \alpha x } $, then : $ y^{(n)} (x) = \displaystyle \sum_{ k=0}^{n} C_{n}^{k} \alpha^k v^{(n-k)} $. So : $ y^{(n)} - y = e^{ \alpha x } $ involves : $ ( \displaystyle \sum_{ k=0}^{n} C_{n}^{k} \alpha^k v^{(n-k)} - v(x) ) e^{ \alpha x } = e^{ \alpha x } $ $\endgroup$ – Bryan261 Mar 14 '15 at 12:04
  • $\begingroup$ @Bryan261, you are welcome. $\endgroup$ – abel Mar 14 '15 at 12:56
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Hint:

Try $y_P=A x e^{\alpha x}$ for some undetermined coefficient $A$. Plug this into your differential equation and find $A$.

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  • $\begingroup$ Thank you very much for your answer , however, the particular solution : $ y_P (x) = Ax e^{ \alpha x} $ is valid only for quadratic equations of type: $ a_2 y '' + a_1 y ' + a_0 y = e^{ \alpha x} $ and not for equations of the nth degree of type : $ \displaystyle \sum_{ k = 0 }^{n} a_k y^{ (k)} = e^{ \alpha x} $, right? Thank you in advance for your help. $\endgroup$ – Bryan261 Mar 14 '15 at 11:39
  • $\begingroup$ It is a trial solution. It could be valid for any linear DE with constant coefficients if it works. A typical way for this kind of problem, when the right hand side is part of the $y_C$, is to multiply the trial solution by $x$. $\endgroup$ – KittyL Mar 14 '15 at 11:45
  • $\begingroup$ If we set $ y (x) = A x e^{ \alpha x} $ , I get by recurrence the following thing : $ y^{ (n)} (x) = A (n \alpha^{ n-1 } + \alpha^n x) e^{ \alpha x} $ . And therefore , $ y^{ (n)} - y = e^{\alpha x} $ involves: $ A (n \alpha^{ n- 1} + ( \alpha^n - 1) x ) e^{\alpha x } = e^{ \alpha x} $. This is ultimately incompatible I think. $\endgroup$ – Bryan261 Mar 14 '15 at 12:01
  • $\begingroup$ Remember $\alpha^n=1$? $\endgroup$ – KittyL Mar 14 '15 at 12:10
  • $\begingroup$ Ah, yes, thank you very much. :-) $\endgroup$ – Bryan261 Mar 14 '15 at 12:51

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