[理工] [工數] [正交曲線]

[asdf322505 ()]

      1
y'=--------- ，通過x-y平面上 (-3,0)點之曲線
    x+y^2


ans; x+y^2+2y+2=-e^y

這提該解怎麼解@@

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◆ From: 122.117.199.34
→ mzhrqoc01:y'=dy/dx=(dx/dy)^-1=(x')^-1
→ mzhrqoc01:=>x'-x=y^2  =>x=c*e^y+(-y^2-2y-2)
→ mzhrqoc01:(-3,0)代入 => c=-1 =>x=-e^y+(-y^2-2y-2)
→ asdf322505:感謝大大的幫忙^^