rogerroger15
contestada

uppose y(t)=9e−5t
y
(
t
)
=
9
e
−
5
t
is a solution of the initial value problem y′+ky=0
y
′
+
k
y
=
0
, y(0)=y0
y
(
0
)
=
y
0
. What are the constants k
k
and y

Respuesta :

LammettHash
If [tex]y(t)=9e^{-5t}[/tex] satisfies the ODE [tex]y'-ky=0[/tex], then

[tex]y'(t)=-45e^{-5t}[/tex]

and substituting into the ODE gives

[tex]-45e^{-5t}-9ke^{-5t}=0\implies k=-5[/tex]

so that the ODE is [tex]y'+5y=0[/tex].

I'm assuming that you're also supposed to determine the initial value [tex]y_0[/tex] given this ODE. You have

[tex]y'+5y=0\implies e^{5t}y'+5e^{5t}y=0\implies(e^{5t}y)'=0\implies e^{5t}y=C\implies y=Ce^{-5t}[/tex]

Given [tex]y(0)=y_0[/tex], you have

[tex]y_0=C[/tex]

but since you already know that [tex]y=9e^{-5t}[/tex], it follows that [tex]C=9[/tex], and in turn that [tex]y_0=9[/tex].

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