Re: [工數]一題微分方程
※ 引述《a016258 (憨仔)》之銘言：
: ※ 引述《peter4108ml (peter4108m)》之銘言：
: : (t^2+2t)x'(t)=2x+2
: : x(1)=1
: : 求x(4)
: dx dt 1 1
: ------- = ----- ( ----- - ----- )
: 2x+2 2 t t+2
: => ln(2x+2) = 1/2 ln [ t / (t+2) ] + C
:
1
: => ln(4) = 1/2 ln ------- + C => C = ln4 - 1/2 ln(1/3)
: 3
:
=> x(4) => ln(2x+2) = 1/2 ln(4/6) + ln4 - 1/2 ln(1/3)
=> ln(2x+2)=ln[4*2^(1/2)]
=> 2x+2=4*2^(1/2) => x=2^(3/2)-1
可以告訴我哪裡錯了嗎?
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