# Fourth Order Homogeneous Linear Equation with real roots and complex roots.

• Feb 22nd 2013, 03:34 AM
bakinbacon
Fourth Order Homogeneous Linear Equation with real roots and complex roots.
the problem is:

y(4)-5y''-36y=0

=> m4-5y2-36=0
=> (m2-9)(m2+4)=0

therefore, m1=3, m2=-3, m3=2i, m4=-2i.

I have learned how to find the general solution for cases with distinct real roots and for cases with complex roots but I am unsure how to I would go about finding the general solution for a linear equation that has both real and complex roots.

Would the general solution be the following

c1e3x+c2e-3x+c3(cosx+(2i)sinx)+c4(cosx+(-2i)sinx)

by using euler's formula?
• Feb 22nd 2013, 04:10 AM
zzephod
Re: Fourth Order Homogeneous Linear Equation with real roots and complex roots.
Quote:

Originally Posted by bakinbacon
the problem is:

y(4)-5y''-36y=0

=> m4-5y2-36=0
=> (m2-9)(m2+4)=0

therefore, m1=3, m2=-3, m3=2i, m4=-2i.

I have learned how to find the general solution for cases with distinct real roots and for cases with complex roots but I am unsure how to I would go about finding the general solution for a linear equation that has both real and complex roots.

Would the general solution be the following

c1e3x+c2e-3x+c3(cosx+(2i)sinx)+c4(cosx+(-2i)sinx)

by using euler's formula?

The general solution is:

$\displaystyle y(x)=\sum_{k=1}^4 A_ke^{\lambda_k x}$

where $\displaystyle \lambda_k, k=1, ... 4$ are the roots of the characteristic equation, irrespective of if they are real imaginary or non-trivially complex.

And for your information: $\displaystyle e^{\pm 2 i x}=\cos(2x) \pm i \sin(2x)$
.