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解微分方程 2ysin2y y0pi01

  令y=P,则y=dP/dx=dP/dy×dy/dx=p×dP/dy,原方程化为:2PdP/dy=sin(2y).

  分离变量:2PdP=sin(2y)dy

  两边积分:P^2=-1/2×cos(2y)+C1,即(y)^2=-1/2×cos(2y)+C1,代入y=π/2,y=1得C1=1/2,所以(y)^2=-1/2×cos(2y)+1/2=(siny)^2

  所以,y=±siny

  由初始条件,y=π/2,y=1>0得y=siny(y=-siny舍去)

  分离变量:cscydy=dx

  两边积分:ln(tan(y/2))=x+lnC,所以,tan(y/2)=Ce^x

  代入x=0,y=π/2得C=1,所以,tan(y/2)=e^x