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

the problem is:

y^{(4)}-5y''-36y=0

=> m^{4}-5y^{2}-36=0

=> (m^{2}-9)(m^{2}+4)=0

therefore, m_{1}=3, m_{2}=-3, m_{3}=2i, m_{4}=-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

c_{1}e^{3}^{x}+c_{2}e^{-3x}+c_{3}(cosx+(2i)sinx)+c_{4}(cosx+(-2i)sinx)

by using euler's formula?

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

=> m^{4}-5y^{2}-36=0

=> (m^{2}-9)(m^{2}+4)=0

therefore, m_{1}=3, m_{2}=-3, m_{3}=2i, m_{4}=-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

c_{1}e^{3}^{x}+c_{2}e^{-3x}+c_{3}(cosx+(2i)sinx)+c_{4}(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: $\displaystyle e^{\pm 2 i x}=\cos(2x) \pm i \sin(2x)$

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