Question

verify that x1(t) = 2e −t − 2e −2t , x2(t) = e −t − 2e −2t solves the system x 0 1 = −2x2, x 0 2 = x1 − 3x2

Answer 1

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Explanation:

First, let’s find the derivatives of x1(t) and x2(t):

x1'(t) = d/dt [2e^(-t) - 2e^(-2t)] = -2e^(-t) + 4e^(-2t)

x2'(t) = d/dt [e^(-t) - 2e^(-2t)] = -e^(-t) + 4e^(-2t)

Now let’s substitute these derivatives into the given system of equations:

x1' = -2x2 becomes -2e^(-t) + 4e^(-2t) = -2(e^(-t) - 2e^(-2t)), which simplifies to -2e^(-t) + 4e^(-2t) = -2e^(-t) + 4e^(-2t).

x2' = x1 - 3x2 becomes -e^(-t) + 4e^(-2t) = (2e^(-t) - 2e^(-2t)) - 3(e^(-t) - 2e^(-2t)), which simplifies to -e^(-t) + 4e^(-2t) = -e^(-t) + 4e^(-2t).

Since both equations are true, we can conclude that x1(t) and x2(t) do indeed solve the given system of equations.