求大神解决不定积分dy/dx=(1-y)^0.92_数学解答

求大神解决不定积分dy/dx=(1-y)^0.92

人气：100 ℃　时间：2020-05-20 18:00:05

解答

dy/(1-y)^0.92=dx
(1-y)^(-0.92)dy=dx
积分
(1-y)^(1-0.92)/(1-0.92)=x+C'
(1-y)^0.08=0.08x+0.08C'=0.08x+C
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