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Higher order derivatives?
given that y = [e^(kt)][cos(pt)], prove y" - 2ky' + (k^2 + p^2)y = 0
HELP !
1 Answer
- kbLv 78 years agoFavorite Answer
Since y = e^(kt) cos(pt), we have by differentiation
y' = ke^(kt) * cos(pt) + e^(kt) * -p sin(pt)
...= e^(kt) [k cos(pt) - p sin(pt)].
y'' = ke^(kt) [k cos(pt) - p sin(pt)] + e^(kt) [-kp sin(pt) - p^2 cos(pt)]
...= e^(kt) [(k^2 - p^2) cos(pt) - 2kp sin(pt)].
Therefore, y" - 2ky' + (k^2 + p^2)y
= e^(kt) [(k^2 - p^2) cos(pt) - 2kp sin(pt)] - 2ke^(kt) [k cos(pt) - p sin(pt)] + (k^2+p^2) e^(kt) cos(pt)
= e^(kt) {[(k^2 - p^2) cos(pt) - 2kp sin(pt)] - 2k [k cos(pt) - p sin(pt)] + (k^2+p^2) cos(pt)}
= e^(kt) {(k^2 - p^2) cos(pt) - 2kp sin(pt) - 2k^2 cos(pt) + 2kp sin(pt) + (k^2+p^2) cos(pt)}
= 0, as required.
I hope this helps!
