Question

解方程dy／dx+(cotx)y=cscx

Answer 1

y'+(cotx)y=cscx.类型：y'+p(x)y=q(x),

(1)现在P(x)=cotx,它的一个原函数是ln(sinx).

又q(x)=cscx,积分∫q(x)e^[ln(sinx)]dx=∫(cscx)(sinx)dx=∫dx=x+C,(C是任意常数)

(2)由解的公式,得y=e^[-ln(sinx)]∫q(x)e^[ln(sinx)]dx=(1/sinx)[x+C]=[x+C]cscx,(C为任意常数).

故y=[x+C]cscx,(C为任意常数) --------------------- (代入原式验证,正确.)