y''(e^x+1)+y'=0的通解

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y''''(e^x+1)+y''=0的通解y''''(e^x+1)+y''=0的通解y''''(e^x+1)+y''=0的通解∵y''''(e^x+1)+y''=0==>(e^x+1)dy''/dx=-y''==>dy''/y''=

y''(e^x+1)+y'=0的通解
y''(e^x+1)+y'=0的通解

y''(e^x+1)+y'=0的通解
∵y''(e^x+1)+y'=0 ==>(e^x+1)dy'/dx=-y'
==>dy'/y'=-dx/(e^x+1)
==>dy'/y'=-e^(-x)dx/(1+e^(-x))
==>dy'/y'=d(1+e^(-x))/(1+e^(-x))
==>ln│y'│=ln(1+e^(-x))+ln│C1│ (C1是积分常数)
==>y'=C1(1+e^(-x))
==>y=C1(x-e^(-x))+C2 (C1是积分常数)
∴原方程的通解是y=C1(x-e^(-x))+C2.