ashleyjsansano4015 ashleyjsansano4015
  • 20-10-2020
  • Mathematics
contestada

Find the general solution of the given higher-order differential equation. y''' + 3y'' + 3y' + y = 0

Respuesta :

raphealnwobi
raphealnwobi raphealnwobi
  • 21-10-2020

Answer:

[tex]y(x)=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}[/tex]

Step-by-step explanation:

We can determine the characteristic equation all we have to do is:

let y' = m, hence y" = m², y'" = m³

Therefore, the higher-order differential equation becomes:

m³ + 3m² + 3m + 1 = 0

(m + 1)(m² + 2m + 1) = 0

(m + 1)(m + 1)(m + 1) = 0

(m + 1)³ = 0

m = -1, -1, -1

Therefore the general solution is given as:

[tex]y(x)=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}[/tex]

[tex]c_1,c_2,c_3\ are\ constants.[/tex]

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