题目

求下列微分方程的通解dy/dx+2xy-4x=0

提问时间：2020-10-29

答案

∵dy/dx+2xy-4x=0
==>dy+2xydx-4xdx=0
==>e^(x^2)dy+2xye^(x^2)dx-4xe^(x^2)dx=0 (等式两端同乘e^(x^2))
==>e^(x^2)dy+yd(e^(x^2))-2d(e^(x^2))=0
==>d(ye^(x^2))-2d(e^(x^2))=0
==>ye^(x^2)-2e^(x^2)=C (C是常数)
==>y=Ce^(-x^2)+2
∴原方程的通解是y=Ce^(-x^2)+2.