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Ejercicios resueltos de sistemas de ecuaciones diferenciales
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Ejercicios resueltos de sistemas de ecuaciones diferenciales
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Solución de sistemas de ecuaciones diferenciales.Un sistema de ecuaciones diferenciales se puede resolver por diferentes métodos algebraicos o por determinantes.
Ejemplo
Resuelva el sistema de ecuaciones diferenciales.
𝑑𝑥 1 2 𝑑𝑦 2 2
= 𝑦− 𝑥 ; = 𝑥− 𝑦 sujetas a 𝑥(0) = 50 y 𝑦(0) = 50
𝑑𝑡 50 25 𝑑𝑡 25 25
Solución:
𝒅𝒙 𝒅𝒚
𝟓𝟎 + 𝟒𝒙 − 𝒚 = 𝟎 ; 𝟐𝟓 𝒙 + 𝟐𝒚 − 𝟐𝒙 = 𝟎
𝒅𝒙 𝒅𝒙
(𝟓𝟎𝑫 + 𝟒)𝒙 − 𝒚 = 𝟎
−𝟐𝒙 + (𝟐𝟓𝑫 + 𝟐)𝒚 = 𝟎
𝟓𝟎𝑫 + 𝟒 −𝟏
∆= | | = (𝟓𝟎𝑫 + 𝟒)(𝟐𝟓𝑫 + 𝟐) − 𝟐 = 𝟏𝟐𝟓𝟎𝑫𝟐 + 𝟐𝟎𝟎𝑫 + 𝟔
−𝟐 𝟐𝟓𝑫 + 𝟐
𝟎 −𝟏 𝟓𝟎𝑫 + 𝟒 𝟎
∆𝟏 = | |=𝟎 ; ∆𝟐 = | |=𝟎
𝟎 𝟐𝟓𝑫 + 𝟐 −𝟐 𝟎
Como ∆𝒙(𝒕) = ∆𝟏 ; ∆𝒙(𝒕) = ∆𝟐
Utilizando la ecuación auxiliar.
−𝟏 −𝟑
𝟏𝟐𝟓𝟎𝒎𝟐 + 𝟐𝟎𝟎𝒎 + 𝟔 = 𝟎 ; (𝟓𝟎𝒎 + 𝟐)(𝟐𝟓𝒎 + 𝟑) = 𝟎 ; 𝒎𝟏 = 𝟐𝟓
; 𝒎𝟐 = 𝟐𝟓
−𝒕 −𝟑𝒕
𝒙(𝒕) = 𝑪𝟏 𝒆𝟐𝟓 + 𝑪𝟐 𝒆 𝟐𝟓 ;
−𝒕 −𝟑𝒕
𝒚(𝒕) = 𝑪𝟑 𝒆 + 𝑪𝟒 𝒆 𝟐𝟓 𝟐𝟓
Ahorra evaluamos las constantes 𝑪𝟑 𝒚 𝑪𝟒 𝒆𝒏 𝒕é𝒓𝒎𝒊𝒏𝒐𝒔 𝒅𝒆 𝑪𝟏 𝒚 𝑪𝟐.
−𝒕 −𝟑𝒕 −𝒕 −𝟑𝒕 −𝒕 −𝟑𝒕
𝒅𝒙 𝑪 𝟑
𝟓𝟎 𝒅𝒙 + 𝟒𝒙 − 𝒚 = 𝟎 ; 𝟓𝟎 (− 𝟐𝟓𝟏 𝒆𝟐𝟓 − 𝟐𝟓 𝑪𝟐 𝒆 𝟐𝟓 ) + 𝟒(𝑪𝟏 𝒆𝟐𝟓 + 𝑪𝟐 𝒆 𝟐𝟓 )= 𝑪𝟑 𝒆𝟐𝟓 + 𝑪𝟒 𝒆 𝟐𝟓
−𝒕 −𝟑𝒕 −𝒕 −𝟑𝒕
2 𝑪𝟏 𝒆𝟐𝟓 − 𝟐𝑪𝟐 𝒆 𝟐𝟓 = 𝑪𝟑 𝒆𝟐𝟓 + 𝑪𝟒 𝒆 𝟐𝟓 ; 𝑪𝟑 = 𝟐𝑪𝟏 , 𝑪𝟒 = −𝟐𝑪𝟐
−𝒕 −𝟑𝒕
𝒙(𝒕) = 𝑪𝟏 𝒆 + 𝑪𝟐 𝒆 𝟐𝟓 𝟐𝟓 ;
−𝒕 −𝟑𝒕
𝒚(𝒕) = 𝟐𝑪𝟏 𝒆 − 𝟐𝑪𝟐 𝒆 𝟐𝟓 𝟐𝟓 ; si 𝒕 = 𝟎 𝒙(𝟎) = 𝟓𝟎 ; 𝒕 = 𝟎 𝒚(𝟎) = 𝟓𝟎
𝟓𝟎 = 𝑪𝟏 + 𝑪𝟐 ;
𝟕𝟓 𝟐𝟓
𝟓𝟎 = 𝟐𝑪𝟏 − 𝟐𝑪𝟐 ; 𝑪𝟏 = 𝑪𝟐 = entonces:
𝟐 𝟐
𝟕𝟓 −𝒕 𝟐𝟓 −𝟑𝒕
𝒙(𝒕) = 𝟐
𝒆𝟓 + 𝟐
𝒆𝟓 ;
−𝒕 −𝟑𝒕
𝒚(𝒕) = 𝟕𝟓𝒆 − 𝟐𝟓𝒆 𝟓 𝟓
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