Question

Find the general solution of the homogeneous differential equation with constant coefficients whose roots of its characteristic equation are given. 1 of multiplicity 2 and -1. y = c_1e^{-t} + c_2e^{-t} + c_3e^t y = c_1e^t + c_2e^t + c_3e^{-t} y = c_1e^t + c_2te^t + c_3e^{-t} y = c_1e^{-t} + c_2te^{-t} + c_3e^t

          Find the general solution of the homogeneous differential equation with constant coefficients whose roots of its characteristic equation are given.

1 of multiplicity 2 and -1.

y = c_1e^{-t} + c_2e^{-t} + c_3e^t
y = c_1e^t + c_2e^t + c_3e^{-t}
y = c_1e^t + c_2te^t + c_3e^{-t}
y = c_1e^{-t} + c_2te^{-t} + c_3e^t

Find the general solution of the homogeneous differential equation with constant coefficients whose roots of its characteristic equation are given.

1 of multiplicity 2 and -1.

y = c1e^-t + c2e^-t + c3e^t
y = c1e^t + c2e^t + c3e^-t
y = c1e^t + c2te^t + c3e^-t
y = c1e^-t + c2te^-t + c3e^t

Added by Marcos V.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

07/10/2023

Step 1

Since the root 1 has multiplicity 2, we will have two linearly independent solutions associated with it: $e^t$ and $te^t$. Show more…

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Find the general solution of the homogeneous differential equation with constant coefficients whose roots of its characteristic equation are given: 1 of multiplicity 2 and -1.