※ 引述《j0958322080 (Tidus)》之銘言：
: 1.mdv/dt = mg - kv^2
: 請問用三角代換後他的上下限要怎麼決定??
: 另外若v>(mg/k)^(1/2)的上下限呢??
mdv/dt = mg - kv^2
dv/dt = g - (k/m)v^2
dv/[g - (k/m)v^2] = dt
dv/[1 - (k/mg)v^2] = gdt
____
dv/[1 - ( √k/mg v )^2] = gdt
____ ____ ____
d ( √k/mg v) / [1 - ( √k/mg v )^2] = ( √kg/m ) dt
-1
∫1/(1-x^2)dx = (1/2) ㏑| (1+x)/(1-x) | + C = tanh x + C
-1 ____ -1 ____ ____ ____
tanh [ √k/mg v(t) ] - tanh [ √k/mg v(0) ] = √kg/m ( t - 0 ) = √kg/m t
-1 ____ ____ -1 ____
tanh [ √k/mg v(t) ] = √kg/m t + tanh [ √k/mg v(0) ]
____ ____ -1 ____
√k/mg v(t) = tanh ( √kg/m t + tanh [ √k/mg v(0) ] )
____ ____ -1 ____
v(t) = √mg/k tanh ( √kg/m t + tanh [ √k/mg v(0) ] )
____
when t → ∞ => v(t) → √mg/k
: 2.dy/dt + 10y = 1,y(1/10)=2/10
: 我的解法：let u(x)=10,∫10dx=10x,e^u(x)=e^10x
: ---> e^10x(y)=∫(e^10x)dx=e^10x/10-->y=1/10
: 不知道哪步錯了
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