微分方程у＇+2у=4χ满足初始条件уχ=0=0的特解为 A、у=2χ+1+e-2χ B、у=2χ-1+e-2χ 试题答案_出自其他系统河南质量工程职业学院高等数学1

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微分方程у＇+2у=4χ满足初始条件уχ=0=0的特解为 A、у=2χ+1+e-2χ B、у=2χ-1+e-2χ C、у=-2χ+1+e-2χ D、у=-2χ-1+e-2χ
答案是:B、у=2χ-1+e-2χ 出自河南质量工程职业学院高等数学1 其他系统河南大学
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