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

**bakinbacon** · February 22nd 2013, 03:34 AM

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

c₁e³^x+c₂e^-3x+c₃(cosx+(2i)sinx)+c₄(cosx+(-2i)sinx)

by using euler's formula?

**zzephod** · February 22nd 2013, 04:10 AM

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

c₁e³^x+c₂e^-3x+c₃(cosx+(2i)sinx)+c₄(cosx+(-2i)sinx)

by using euler's formula?

The general solution is:

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

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

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

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