| u = u(t) = x1(t), v = v(t) = x2(t) とおきます。
u' = -5u+4v・・・・・(1) v' = -9u+7v+t(e^t)・・・・・(2)
(1)より、 v = (1/4)(u'+5u)・・・・・(3) v' = (1/4)(u''+5u')・・・・・(4)
(3)(4)を(2)に代入して、 (1/4)(u''+5u') = -9u+7(1/4)(u'+5u)+t(e^t) ⇒ u''+5u' = -36u+7(u'+5u)+4t(e^t) ⇒ u''-2u'+u = 4t(e^t) ⇒ (u''-u')-(u'-u) = 4t(e^t) ⇒ {(u'-u)(e^(-t))}' = 4t ⇒ (u'-u)(e^(-t)) = 2t^2+C (Cは積分定数) ⇒ {u(e^(-t))}' = 2t^2+C ⇒ u(e^(-t)) = (2/3)t^3+Ct+D (Dは積分定数) ⇒ u = (e^t){(2/3)t^3+Ct+D}
検算 u' = (e^t){(2/3)t^3+Ct+D}+(e^t){2t^2+C} = (e^t){(2/3)t^3+2t^2+Ct+C+D} u'' = (e^t){(2/3)t^3+2t^2+Ct+C+D}+(e^t){2t^2+4t+C} = (e^t){(2/3)t^3+4t^2+(C+4)t+2C+D} {u''-2u'+u}(e^(-t)) = {(2/3)t^3+Ct+D}-2{(2/3)t^3+2t^2+Ct+C+D}+{(2/3)t^3+4t^2+(C+4)t+2C+D} = 4t OK!
上記結果を(3)に代入して、 v = (1/4)(e^t){{(2/3)t^3+2t^2+Ct+C+D}+5{(2/3)t^3+Ct+D}} = (1/4)(e^t){(12/3)t^3+2t^2+6Ct+C+6D} = (e^t){t^3+(1/2)t^2+(3/2)Ct+(C+6D)/4}
検算 v' = (e^t){t^3+(1/2)t^2+(3/2)Ct+(C+6D)/4}+(e^t){3t^2+t+(3/2)C} = (e^t){t^3+(7/2)t^2+(3C/2+1)t+(7C+6D)/4}
-9u+7v+t(e^t) = (e^t){-9{(2/3)t^3+Ct+D}+7{t^3+(1/2)t^2+(3/2)Ct+(C+6D)/4}+t} = (e^t){-6t^3-9Ct-9D+7t^3+(7/2)t^2+(21/2)Ct+(7C+42D)/4+t} = (e^t){t^3+(7/2)t^2+(3C/2+1)t+(7C+6D)/4} OK!
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