# Linear differential equation to solve.

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 $(E)$.

$\bf hint:$ 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 $$v = \frac 1{3\alpha^2}x,$$

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

• $v=v(x)$, right ? – Bryan261 Mar 14 '15 at 11:41
• @Bryan261,yes. $v$ is a function of $x.$ – abel Mar 14 '15 at 11:42
• 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 : $( \displaystyle \sum_{ k=0}^{n} C_{n}^{k} \alpha^k v^{(n-k)} - v(x) ) e^{ \alpha x } = e^{ \alpha x }$ – Bryan261 Mar 14 '15 at 12:04
• @Bryan261, you are welcome. – abel Mar 14 '15 at 12:56

Hint:

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

• 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. – Bryan261 Mar 14 '15 at 11:39
• 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 $x$. – KittyL Mar 14 '15 at 11:45
• 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. – Bryan261 Mar 14 '15 at 12:01
• Remember $\alpha^n=1$? – KittyL Mar 14 '15 at 12:10
• Ah, yes, thank you very much. :-) – Bryan261 Mar 14 '15 at 12:51