Question

一阶线性微分方程dy/dx+P(x)y=Q(x)的通解公式怎么理解？

Answer 1

一阶线性微分方程dy/dx+P(x)y=Q(x)的通解公式应用“常数变易法”求解。
解：∵由齐次方程dy/dx+P(x)y=0
==>dy/dx=-P(x)y
==>dy/y=-P(x)dx
==>ln│y│=-∫P(x)dx+ln│C│
(C是积分常数)
==>y=Ce^(-∫P(x)dx)
∴此齐次方程的通解是y=Ce^(-∫P(x)dx)
于是，根据常数变易法，设一阶线性微分方程dy/dx+P(x)y=Q(x)的解为
y=C(x)e^(-∫P(x)dx)
(C(x)是关于x的函数)
代入dy/dx+P(x)y=Q(x)，化简整理得
C'(x)e^(-∫P(x)dx)=Q(x)
==>C'(x)=Q(x)e^(∫P(x)dx)
==>C(x)=∫Q(x)e^(∫P(x)dx)dx+C
(C是积分常数)
==>y=C(x)e^(-∫P(x)dx)=[∫Q(x)e^(∫P(x)dx)dx+C]e^(-∫P(x)dx)
故一阶线性微分方程dy/dx+P(x)y=Q(x)的通解公式是
y=[∫Q(x)e^(∫P(x)dx)dx+C]e^(-∫P(x)dx)
(C是积分常数)。