微分方程的特解本人高数奇差 麻烦大侠帮我步骤写详细清楚点哈都是一阶线性微分方程 用那个dy/dx+P(x)y=Q(x)的

1.∵齐次方程y'+y/x=0 ==>dy/y=-dx/x
==>ln│y│=ln│C│-ln│x│ (C是积分常数)
==>y=C/x
∴设原方程的通解是y=C(x)/x (C(x)是关于x的函数)
∵y'=[C'(x)x-C(x)]/x²
代入原方程,得[C'(x)x-C(x)]/x²+C(x)/x²=sinx/x
==>C'(x)=sinx
==>C(x)=-cosx+C (C是积分常数)
∴原方程的通解是y=(C-cosx)/x
∵当x=π时y=1 ==>(C+1)/π=1 ==>C=π-1
∴满足条件的解是y=(π-1-cosx)/x
2.∵齐次方程y'+ycotx=0 ==>dy/y=-cosxdx/sinx
==>dy/y=-d(sinx)/sinx
==>ln│y│=ln│C│-ln│sinx│ (C是积分常数)
==>y=C/sinx
∴设原方程的通解是y=C(x)/sinx (C(x)是关于x的函数)
∵y'=[C'(x)sinx-C(x)cosx]/sin²x
代入原方程,得[C'(x)sinx-C(x)cosx]/sin²x+[C(x)/sinx]cotx=5e^cosx
==>C'(x)=5sinx*e^cosx
==>C(x)=-5∫e^cosxd(cosx)
==>C(x)=-5e^cosx+C (C是积分常数)
∴原方程的通解是y=(C-5e^cosx)/sinx
∵当x=π/2时,y=-4 ==>C-5=-4 ==>C=1
∴满足条件的解是y=(1-5e^cosx)/sinx
==>ysinx+5e^cosx=1