解ï¼å为
y'+tanx*y=1/cosx
P(x)=tanx
Q(x)=1/cosx
æ
â«-P(x)dx=â«-tanxdx=lncosx
e^[â«-P(x)dx]=cosx
â«Q(x)*e^[â«P(x)dx]dx=â«1/cosx*e^(-lncosx)dx=â«1/cos²xdx=tanx
æ
é解为
y=e^â«-P(x)dx*{â«Q(x)*e^[â«P(x)dx]dx+C}
=cosx*[tanx+C]
=sinx+Ccosx
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