Given the equation
$$
4 y^{\prime \prime}+3 y^{\prime}-y=0
$$
and its solution $y=e^{\lambda t}$, what are the values of $\lambda$ ?
(USA Rice University Mathematics Tournament)
$$
\left[-1 \vee \frac{1}{4}\right]
$$
Given the equation
$$
4 y^{\prime \prime}+3 y^{\prime}-y=0
$$
and its solution $y=e^{\lambda t}$, what are the values of $\lambda$ ?
(USA Rice University Mathematics Tournament)
$$
\left[-1 \vee \frac{1}{4}\right]
$$
90
punti
Livello 1
y = e^(λ·t)
y'=λ·e^(t·λ)
y''=λ^2·e^(t·λ)
Quindi deve essere:
4·(λ^2·e^(t·λ)) + 3·(λ·e^(t·λ)) - e^(λ·t) = 0
e^(t·λ)·(4·λ^2 + 3·λ - 1) = 0
4·λ^2 + 3·λ - 1 = 0
risolvo ed ottengo:
λ = 1/4 ∨ λ = -1
90143
punti
Genius II
@lucianop grazie mille
Di nulla. Buona giornata.