2nd order homogeneous equations complex root

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SUMMARY

The discussion centers on solving the second-order homogeneous differential equation y'' - 2y' + 5y = 0, which yields complex conjugate roots. The correct homogeneous solutions are confirmed as sin(ωt)e^(σt) and cos(ωt)e^(σt), where the roots take the form r = σ ± ωi, with 'i' representing the imaginary unit. The initial conditions provided are y(0) = 1 and y'(0) = 1, which are essential for determining specific constants in the solution.

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Homework Statement


y'' -2y' +5y =0 , y(0)=1, y'(0)=1
you get a complex root conjugate.

Homework Equations


y=e^(rt)
y'=re^(rt)
y''=r^2 * e^(rt)

The Attempt at a Solution


I have in my notes sin(omega*t)e^(sigma *t), cos(omega *t)e^(sigma).
I don't think i took down the notes properly, can anyone care confirm if this is right or wrong?
 
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Yes your homogeneous solutions are sin(ωt)eσt and cos(ωt)eσt when the roots are in the form r=σ ± ωi with 'i' being the imaginary unit.
 

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