Respuesta :
Answer:
[tex]y(t) = C_1 e^{-9t}+C_2te^{-9t}[/tex]
Step-by-step explanation:
Consider the differential equation [tex]\frac{d^2y}{dt^2} + 18 \frac{dy}{dt}+81y=0[/tex]. Let us propose a solution of the form [tex] y = A e^{rt}[/tex] where A is a non-zero constant. Then, the derivatives are calculated as follows: [tex]\frac{dy}{dt} = Ar e^{rt},\frac{d^2y}{dt^2}= Ar^2e^{rt}[/tex]. Replacing those functions in the original equation, we obtain the following.
[tex]Ar^2e^{rt}+18Are^{rt}+81Ae^{rt}=0= Ae^{rt}(r^2+18r+81)=0[/tex]
Given that A is non-zero and that the exponential function is non-zero, the following equation must be hold
[tex]r^2+18r+81 = 0=(r+9)^2[/tex]
Then, the only solution is r=-9. Then, one of the solutions is the function [text]y_1(t)=C_1 e^{-9t}[/tex]. Since the differential equation is a second degree equation, another solution must be given. Note that the second solution must be linear independent of the first found solution. This is achieved by multiplying the first solution by t i.e [tex]y_2(t) = C_2 te^{-9t}[/tex]. Then, the complete solution is of the form [tex]y(t) = C_1 e^{-9t}+C_2te^{-9t}[/tex]
The correct question is:
Find the general solution to the homogeneous differential equation d²y/dt² + 18dy/dt + 81y = 0.
The solution can be written in the form:
y = C1f1(t) + C2f2(t)
with f1(t) = ? and f2(t) = ?
Answer:
f1(t) = e^(9t)
f2(t) = te^(9t)
Step-by-step explanation:
Give the homogeneous differential equation:
d²y/dt² + 18dy/dt + 81y = 0
To solve this, let us write and solve the auxiliary equation:
m² + 18m + 81 = 0
m² + 9m + 9m + 81 = 0
(m² + 9m) + (9m + 81) = 0
m(m + 9) + 9(m + 9) = 0
(m + 9)(m + 9) = 0
m = 9 twice.
For a repeated root, we have a solution of the form:
y = (C1 + C2t)e^(mt)
Our solution here is therefore,
y = (C1 + C2t)e^(9t)
= C1e^(9t) + C2te^(9t)
Comparing with
y = C1f1(t) + C2f2(t)
We have
f1(t) = e^(9t)
f2(t) = te^(9t)
