y''+(y')²=e^(-y)/2的通解

Question

y''+(y')²=e^(-y)/2的通解

Answer 1

∵y''+(y')^2=e^(-y)/2
==>y'dy'/dy+(y')^2=e^(-y)/2 (y"=y'dy'/dy)
==>2y'dy'+2(y')^2dy=e^(-y)dy
==>2y'e^(2y)dy'+2(y')^2e^(2y)dy=e^ydy (等式两端同乘e^(2y))
==>e^(2y)d((y')^2)+(y')^2d(e^(2y))=d(e^y)
==>d((y')^2e^(2y))=d(e^y)
==>(y')^2e^(2y)=e^y+C0 (C0是任意常数)
==>y'e^y=±√(e^y+C0) (等式两端开平方)
==>e^ydy/dx=±√(e^y+C0)
==>e^ydy=±√(e^y+C0)dx
==>e^ydy/√(e^y+C0)=±dx
==>d(e^y+C0)/√(e^y+C0)=±dx
==>√(e^y+C0)=C1±x/2 (C1是任意常数)
==>e^y+C0=(C1±x/2)^2 (等式两端取平方)
==>e^y=(C1±x/2)^2-C0
==>e^y=x^2/4+C1x+(C1^2-C0)
==>e^y=x^2/4+C1x+C2 (令C2=C1^2-C0,则C2也是任意常数)
∴原方程的通解是e^y=x^2/4+C1x+C2.