Question

常微分方程，求解非齐次线性方程的初值问题！

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Answer 1

(1) dx/dt + 2x/t = 1, 一阶线性微分方程
x = e^(-∫2dt/t) [∫1e^(∫2dt/t)dt + C1]
= (1/t^2) (∫t^2dt + C1) = (1/t^2) (t^3/3 + C1)
= t/3 + C1/t^2, x(1) = 1/3 代入得 C1 = 0
x = t/3。
dy/dt = t/3 + y - 1 + 2/3,
dy/dt - y = (t-1)/3
y = e^(∫dt) [(1/3)∫(t-1)e^(-∫dt)dt + C2]
= e^t [(1/3)∫(t-1)e^(-t)dt + C2]
= e^t [-(1/3)∫(t-1)de^(-t) + C2]
= e^t [-(1/3)(t-1)e^(-t) + (1/3)∫e^(-t)dt + C2]
= e^t [-(1/3)(t-1)e^(-t) - (1/3)e^(-t) + C2]
= e^t [-(1/3)te^(-t) + C2] = -t/3 + C2e^t
y(1) = -1/3 代入得 C2 = 0, y = -t/3