- #1
jamesbob
- 63
- 0
[tex] \ddot{x} + 10\dot{x} + 34x = 50e^{-t} [/tex]
Im stuck on getting the homogeneous equation. I am used to equations with just first derivatives, not second derivatives as well. Usually, i'd set the left hand side equal to zero and say:
[tex] set \left x = Ae^{\lambda{t}} \left and \left so \left \dot{x} = A \lambda e ^{\lambda{t}} [/tex]
id then put in the coefficients and get something like LAMBDA = -2.
But when there's a second derivative i have an extra lamba squared expression and therefore a polynomial that is equal to zero. This results in 2 solutions - is that okay for the homogeneous equation? Il work thru this question as i would:
[tex] \ddot{x} + 10\dot{x} + 34x = 50e^{-t} [/tex]
[tex] Homogeneous: \ddot{x} + 10\dot{x} + 34x = 0 [/tex]
[tex] set: \left x = Ae^{\lambda{t}} \left and \left \dot{x} = A \lambda e^{\lambda{t}} \left and \left \ddot{x} = A \lambda^2 e^{\lambda{t}} [/tex]
putting in the coef's we have:
[tex] A \lambda^2 e^{\lambda{t}} + 10A \lambda e^{\lambda{t}} + 34Ae^{\lambda{t}} = 0 [/tex]
this simplifies to:
[tex] \lambda^2 + 10\lambda + 34 = 0 [/tex]
Do i now perform the quadratic formula to get 2 results for LAMBDA thus giving me something along the lines of
[tex] Homogeneous eq = Ae^{xt} + Ae^{yt} [/tex] ?
Im stuck on getting the homogeneous equation. I am used to equations with just first derivatives, not second derivatives as well. Usually, i'd set the left hand side equal to zero and say:
[tex] set \left x = Ae^{\lambda{t}} \left and \left so \left \dot{x} = A \lambda e ^{\lambda{t}} [/tex]
id then put in the coefficients and get something like LAMBDA = -2.
But when there's a second derivative i have an extra lamba squared expression and therefore a polynomial that is equal to zero. This results in 2 solutions - is that okay for the homogeneous equation? Il work thru this question as i would:
[tex] \ddot{x} + 10\dot{x} + 34x = 50e^{-t} [/tex]
[tex] Homogeneous: \ddot{x} + 10\dot{x} + 34x = 0 [/tex]
[tex] set: \left x = Ae^{\lambda{t}} \left and \left \dot{x} = A \lambda e^{\lambda{t}} \left and \left \ddot{x} = A \lambda^2 e^{\lambda{t}} [/tex]
putting in the coef's we have:
[tex] A \lambda^2 e^{\lambda{t}} + 10A \lambda e^{\lambda{t}} + 34Ae^{\lambda{t}} = 0 [/tex]
this simplifies to:
[tex] \lambda^2 + 10\lambda + 34 = 0 [/tex]
Do i now perform the quadratic formula to get 2 results for LAMBDA thus giving me something along the lines of
[tex] Homogeneous eq = Ae^{xt} + Ae^{yt} [/tex] ?
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