# Solving $y''''-y'=0$

This DE has characteristic equation $\lambda^{4}-\lambda=0$.

$\lambda(\lambda^{3}-1)=0$

$\lambda_1=0,\lambda_2=\lambda_3=\lambda_4=1$

$y=C_{1}\,y_{1}+C_{2}\,y_{2}+C_{3}\,y_{3}+C_{4}\,y_{4}$, where $y_{1}=e^{0x}$, $y_{2}=e^{1x}$, $y_{3}=x\,e^{1x}$,$y_{4}=x^{2}\,e^{1x}$. So solution should be $y=C_{1}+C_{2}\,e^{x}+C_{3}\,x\,e^{x}+C_{4}\,x^{2}\,e^{x}$, but $y''''-y'=0$ is not satisfied. Where I made mistake?

• Who are the solutions of $\lambda^3-1=0$ ? Aug 14 '15 at 13:24
• I totally forgot on complex roots :)
– etf
Aug 14 '15 at 13:32

HINT: $$\lambda^3-1=0$$ $$(re^{i\theta})^3=1\times e^{i2k\pi}$$

$$r=1 , \theta=\frac{2k\pi}{3}$$ For $k=0,1$ and $2$

• I totally forgot on complex roots :) thanks a lot!
– etf
Aug 14 '15 at 13:33

$\lambda^3-1=(\lambda -1)(\lambda ^2+\lambda+1)$

$\lambda ^2+\lambda+1=0\implies \lambda=\frac{-1\pm \sqrt{3}i}{2}$

• I totally forgot on complex roots :) thanks a lot!
– etf
Aug 14 '15 at 13:33

There are three distinct complex solutions to the equation $\lambda^3-1=0$. You've correctly identified one of them as being $1$; the other two are the primitive third roots of unity, namely $e^{2\pi i/3}$ and $e^{4\pi i/3}$.

\begin{align} 0 &= \lambda^{4} - \lambda = \lambda \, (\lambda^{3} - 1) \\ &= \lambda \, (\lambda - 1) \, (\lambda^{2} + \lambda + 1) \\ &= \lambda \, (\lambda - 1) \, \left(\lambda - \frac{-1+\sqrt{3} i}{2}\right) \, \left(\lambda - \frac{-1-\sqrt{3} i}{2}\right) \end{align} The roots are $$\lambda \in \left\{ 0, 1, - \frac{1 + \sqrt{3} i}{2}, - \frac{1 - \sqrt{3} i}{2} \right\}$$

• I totally forgot on complex roots :) thanks a lot!
– etf
Aug 14 '15 at 13:32