Re: [工數] ode
※ 引述《qq0212q (劉小妍)》之銘言:
: 2
: (x+2) y''-(x+2)y'+y=3x+4
: 我是令z=x+2
: 2
: 原式改寫成:z y''-zy'+y=3z+2
這裡應該是 3z-2
: 先解得yh=c1z+c2z(lnz)
: 但是接下來要用參數變異法卻一直積不出來
: 不知道那裡錯了....
: 2
: ans:y=c1(x+2)+c2(x+2)ln(x+2)+1.5(x+2)[ln(x+2)] -2
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az + b
我猜原 po 應該是卡在型如 ∫ ────*ln(z) dz 的不定積分
z^2
乍看之下好像會沒辦法算出來,不過實際下去算用分佈積分就可以了:
az + b b a*ln(z) b
∫ ────*ln(z) dz = [ a*ln(z) - ── ]*ln(z) - ∫ ──── - ── dz
z^2 z z z^2
b a*[ln(z)]^2 b
= [ a*ln(z) - ── ]*ln(z) - ────── - ── + C
z 2 z
az*[ln(z)]^2 - 2b*ln(z) - 2b
= ────────────── + C
2z
後續你應該會算了
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<另解>
(z^2)y''-zy'+y = 3z-2
=> [z*(y/z)']' = 3/z - 2/z^2
=> (y/z)' = 3*ln(z)/z + 2/z^2 + c1/z
=> y = (3/2)z[ln(z)]^2 - 2 + c1*zln(z) + c2*z
= (3/2)(x+2)[ln(x+2)]^2 - 2 + c1*(x+2)ln(x+2) + c2*(x+2)
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