✔ 最佳答案
問題:
若 dy/dx = 1/(y - 3),請問要如何求出 d²y/dx² ?
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螞蟻雄兵 大師的解答:
dy/dx=1/(y-3)
d^2 y/dx^2
=d[1/(y-3)]/dx
={d[1/(y-3)]/d(y-3)}*[d(y-3)/dx]
=[-1/(y-3)^2]*[1/y-3]]
=-1/(y-3)^3
or
y’=1/(y-3)
yy’-3y’=1
d(yy’-3y’)/dx=0
yy”+(y’)^2-3y”=0
(y-3)y”=-(y’)^2
(y-3)y”=-1/(y-3)^2
y”=-1/(y-3)^3
d^2 y/dx^2=-1/(y-3)^3
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解答:
dy/dx = 1/(y - 3)
(y - 3) dy/dx = 1
d[ (y - 3) dy/dx ]/dx = d(1)/dx
(dy/dx) d(y - 3)/dx + (y - 3) d²y/dx² = 0
(dy/dx)² + (y - 3) d²y/dx² = 0
1/(y - 3)² + (y - 3) d²y/dx² = 0
d²y/dx² = -1/(y - 3)³
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20160712
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