# second order differential equation

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• April 26th 2010, 06:32 AM
bobey
second order differential equation
show that y1(x) = e^(2+i)x and y2(x) = e^(2-i)x, i=sqrt(-1) are two linearly independent functions

hence obtain a second order linear differential equation with constant coefficients each that y1(x) and y2(x) are its two fundamental solutions.

my attempt :

for the first part, I use the definition of wroskian = y1y2'-y2y1' and show it not equal to zero... ok

the second part, I don't know how to do it...

how to get the second order differential equation????

is that setting : (r+2+i)(r+2-i) to get the auxillary equation??? is that possible??? can someone show me to solve this problem??? (Crying)
• April 26th 2010, 08:21 AM
chisigma
Suppose that $s_{1} = 2 + i$ and $s_{2} = 2 - i$ are solution of a second order equation is s. In that case the equation is...

$(s-s_{1})\cdot (s-s_{2})= s^{2} - 4\cdot s + 5 = 0$ (1)

... so that the second order ODE is...

$y^{''} - 4\cdot y^{'} + 5 \cdot y =0$ (2)

Kind regards

$\chi$ $\sigma$