求微分方程y#39;+2xy=x,y(0)=-2的特解
<p>问题：求微分方程y#39;+2xy=x,y(0)=-2的特解<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">江博的回答：<div class="content-b">网友采纳 y'+2xy=xe^(-x)y'+2xy=0y'=-2xydy/y=-2xdxy=C0e^(-x^2)设y=c0(x)e^(-x^2)C0'e^(-x^2)=xe^(-x)dC0=xe^(x^2-x)dx∫xe^(x^2-x)dx=(1/2)∫(2x)e^(x^2-x)dx=(1/2)∫e^(x^2)d(x^2)/e^x=(1/2)∫de^(x^2)/e^x=(1/2)∫d(e^x^2...
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