问题描述: 微分方程(x2-1)dy+(2xy-cosx)dx=0,y|x=0=1的特解为?答案是y=sinx-1/x2-1 1个回答 分类:数学 2014-10-28 问题解答: 我来补答 (x2-1)dy+(2xy-cosx)dx=0dy/dx+2x/(x^2-1)*y=cosx/(x^2-1)这是个一阶非齐次微分方程通解为:y=ce^(-∫P(x)dx)+∫f(x)e^(∫P(x)dx)dx*e^(-∫P(x)dx)这里P(x)=2x/(x^2-1),f(x)=cosx/(x^2-1)显然∫P(x)dx=∫2x/(x^2-1)dx=∫dx^2/x^2-1=ln(x^2-1)所以∫f(x)e^(∫P(x)dx)dx=∫cosx/(x^2-1)*e^[ln(x^2-1)]dx=∫cosxdx=sinx所以通解为y=c/(x^2-1)+sinx/(x^2-1)当x=0时y=1显然有c=-1答案应该加括号解应该是y=-1/(x^2-1)+sinx/(x^2-1) 展开全文阅读