# Higher Order Complex Roots

• Jul 15th 2007, 06:01 PM
fudawala
Higher Order Complex Roots
Find the general solution to the given differential equation:

$
y^{(4)} - 8 y' = 0
$

$
y''' + 5 y'' + 6 y' + 2 y = 0
$

$
4 y''' + y' + 5 y = 0
$
• Jul 15th 2007, 07:27 PM
topsquark
Quote:

Originally Posted by fudawala
Find the general solution to the given differential equation:

$
y^{(4)} - 8 y' = 0
$

The characteristic equation is
$m^4 - 8 = 0$

$(m^2 + \sqrt{8})(m^2 - \sqrt{8}) = 0$

$(m^2 + 2 \sqrt{2})(m^2 - 2 \sqrt{2}) = 0$

The first factor doesn't factorize further over the real numbers, but we can factor it over the complex numbers:
$m^2 + 2sqrt{2} = \left ( m + i \sqrt{2{\sqrt{2}}} \right ) \left ( m - i \sqrt{2{\sqrt{2}}} \right )$

and, of course
$m^2 - 2sqrt{2} = \left ( m + \sqrt{2{\sqrt{2}}} \right ) \left ( m - \sqrt{2{\sqrt{2}}} \right )$

So
$m^4 - 8 = \left ( m + i \sqrt{2{\sqrt{2}}} \right ) \left ( m - i \sqrt{2{\sqrt{2}}} \right )
\left ( m + \sqrt{2{\sqrt{2}}} \right ) \left ( m - \sqrt{2{\sqrt{2}}} \right ) = 0$

Thus the solution to the differential equation will be of the form:
$y(x) = Ae^{ix \sqrt{2{\sqrt{2}}}} + Be^{-i x \sqrt{2{\sqrt{2}}}} + Ce^{x \sqrt{2{\sqrt{2}}}} + De^{-x \sqrt{2{\sqrt{2}}}}$

or if you prefer:
$y(x) = A^{\prime}~cos \left ( x \sqrt{2{\sqrt{2}}} \right ) + B^{\prime}~sin \left ( x \sqrt{2{\sqrt{2}}} \right ) + Ce^{x \sqrt{2{\sqrt{2}}}} + De^{-x \sqrt{2{\sqrt{2}}}}$

-Dan
• Jul 15th 2007, 07:33 PM
topsquark
Quote:

Originally Posted by fudawala
Find the general solution to the given differential equation:

$
y''' + 5 y'' + 6 y' + 2 y = 0
$

The characteristic equation is
$m^3 + 5m^2 + 6m + 2 = 0$

The rational roots theorem says that any rational zeros of this polynomial will be of the form $\pm 1, \pm 2$. Checking these shows that m = -1 is a solution. Thus:
$m^3 + 5m^2 + 6m + 2 = (m + 1)(m^2 + 4m + 2) = 0$

We use the quadratic formula to get factors of the quadratic factor, giving:
$m^3 + 5m^2 + 6m + 2 = (m + 1)(m^2 + 4m + 2) = (m + 1)(x + 2 + \sqrt{2})(x + 2 - \sqrt{2}) = 0$

which has solutions $m = -1, -2 + \sqrt{2}, -2 - \sqrt{2}$.

So the solution of the differential equation will be of the form:
$y(x) = Ae^{-x} + Be^{(-2 + \sqrt{2})x} + Ce^{(-2 - \sqrt{2})x}$

-Dan
• Jul 15th 2007, 07:38 PM
topsquark
Quote:

Originally Posted by fudawala
Find the general solution to the given differential equation:
$
4 y''' + y' + 5 y = 0
$

The characteristic equation is
$4m^3 + m +5 = 0$

The possible rational zeros of this polynomial are:
$m = \pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}$

Checking these I see that $m = -1$ is the only rational solution. Thus:
$4m^3 + m +5 = (m + 1)(4m^2 - 4m + 5) = (m + 1) \left ( m -\frac{1}{2} + i \right ) \left ( m -\frac{1}{2} - i \right ) = 0$

You can take it from here.

-Dan