Find the general solution to the given differential equation:
$\displaystyle
y^{(4)} - 8 y' = 0
$
$\displaystyle
y''' + 5 y'' + 6 y' + 2 y = 0
$
$\displaystyle
4 y''' + y' + 5 y = 0
$
The characteristic equation is
$\displaystyle m^4 - 8 = 0$
$\displaystyle (m^2 + \sqrt{8})(m^2 - \sqrt{8}) = 0$
$\displaystyle (m^2 + 2 \sqrt{2})(m^2 - 2 \sqrt{2}) = 0$
The first factor doesn't factorize further over the real numbers, but we can factor it over the complex numbers:
$\displaystyle m^2 + 2sqrt{2} = \left ( m + i \sqrt{2{\sqrt{2}}} \right ) \left ( m - i \sqrt{2{\sqrt{2}}} \right )$
and, of course
$\displaystyle m^2 - 2sqrt{2} = \left ( m + \sqrt{2{\sqrt{2}}} \right ) \left ( m - \sqrt{2{\sqrt{2}}} \right ) $
So
$\displaystyle m^4 - 8 = \left ( m + i \sqrt{2{\sqrt{2}}} \right ) \left ( m - i \sqrt{2{\sqrt{2}}} \right )
\left ( m + \sqrt{2{\sqrt{2}}} \right ) \left ( m - \sqrt{2{\sqrt{2}}} \right ) = 0$
Thus the solution to the differential equation will be of the form:
$\displaystyle y(x) = Ae^{ix \sqrt{2{\sqrt{2}}}} + Be^{-i x \sqrt{2{\sqrt{2}}}} + Ce^{x \sqrt{2{\sqrt{2}}}} + De^{-x \sqrt{2{\sqrt{2}}}}$
or if you prefer:
$\displaystyle y(x) = A^{\prime}~cos \left ( x \sqrt{2{\sqrt{2}}} \right ) + B^{\prime}~sin \left ( x \sqrt{2{\sqrt{2}}} \right ) + Ce^{x \sqrt{2{\sqrt{2}}}} + De^{-x \sqrt{2{\sqrt{2}}}}$
-Dan
The characteristic equation is
$\displaystyle m^3 + 5m^2 + 6m + 2 = 0$
The rational roots theorem says that any rational zeros of this polynomial will be of the form $\displaystyle \pm 1, \pm 2$. Checking these shows that m = -1 is a solution. Thus:
$\displaystyle m^3 + 5m^2 + 6m + 2 = (m + 1)(m^2 + 4m + 2) = 0$
We use the quadratic formula to get factors of the quadratic factor, giving:
$\displaystyle m^3 + 5m^2 + 6m + 2 = (m + 1)(m^2 + 4m + 2) = (m + 1)(x + 2 + \sqrt{2})(x + 2 - \sqrt{2}) = 0$
which has solutions $\displaystyle m = -1, -2 + \sqrt{2}, -2 - \sqrt{2}$.
So the solution of the differential equation will be of the form:
$\displaystyle y(x) = Ae^{-x} + Be^{(-2 + \sqrt{2})x} + Ce^{(-2 - \sqrt{2})x}$
-Dan
The characteristic equation is
$\displaystyle 4m^3 + m +5 = 0$
The possible rational zeros of this polynomial are:
$\displaystyle m = \pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}$
Checking these I see that $\displaystyle m = -1$ is the only rational solution. Thus:
$\displaystyle 4m^3 + m +5 = (m + 1)(4m^2 - 4m + 5) = (m + 1) \left ( m -\frac{1}{2} + i \right ) \left ( m -\frac{1}{2} - i \right ) = 0$
You can take it from here.
-Dan