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師兄師姐們,幫小弟看看這個 Laplace變換后的二階常微分方程怎么解啊 已有3人參與
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首先,這不是二階常微分方程,“!北硎緔'',y',y的系數(shù)都是常數(shù),形式為ay''+by'+cy=0,這類問題可以用特征方程求解。你的問題是二階非線性微分方程。 我想了下也不知道怎么解,用matlab算: >> syms a s y x Y=dsolve('a*D2y-Dy/x-s*y=0','x') Y = C3*x^((a + 1)/(2*a))*besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)) + (C2*x^((a + 1)/(2*a))*besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/(-1)^((a + 1)/(2*a)) >> pretty(Y) a + 1 ----- / 1/2 \ 2 a | a + 1 s x | a + 1 C2 x besseli| - -----, ------ | ----- / 1/2 \ | 2 a 1/2 | 2 a | a + 1 s x | \ a / C3 x besselk| - -----, - ------ | + ------------------------------------ | 2 a 1/2 | a + 1 \ a / ----- 2 a (-1) >> 解中由用到bessel函數(shù) |
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但是,反過來 ans=a*diff(Y,x,2)-diff(Y,x)/x-s*Y ans = a*(C3*x^((a + 1)/(2*a))*((s^(1/2)*((besselk(1 - (a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*((a + 1)/(2*a) - 1))/x + (s^(1/2)*besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)))/a^(1/2)))/a^(1/2) + (besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x^2) - ((a + 1)*((s^(1/2)*besselk(1 - (a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)))/a^(1/2) - (besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x)))/(2*a*x)) + (C2*x^((a + 1)/(2*a))*((s^(1/2)*((besseli(1 - (a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*((a + 1)/(2*a) - 1))/x + (s^(1/2)*besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/a^(1/2)))/a^(1/2) + (besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x^2) - ((a + 1)*((s^(1/2)*besseli(1 - (a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/a^(1/2) - (besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x)))/(2*a*x)))/(-1)^((a + 1)/(2*a)) + (C3*x^((a + 1)/(2*a) - 1)*(a + 1)*((s^(1/2)*besselk(1 - (a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)))/a^(1/2) - (besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x)))/a + (C3*x^((a + 1)/(2*a) - 2)*((a + 1)/(2*a) - 1)*besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a) + (C2*x^((a + 1)/(2*a) - 1)*(a + 1)*((s^(1/2)*besseli(1 - (a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/a^(1/2) - (besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x)))/((-1)^((a + 1)/(2*a))*a) + (C2*x^((a + 1)/(2*a) - 2)*((a + 1)/(2*a) - 1)*besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*(a + 1))/(2*(-1)^((a + 1)/(2*a))*a)) - (C3*x^((a + 1)/(2*a))*((s^(1/2)*besselk(1 - (a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)))/a^(1/2) - (besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x)) + (C2*x^((a + 1)/(2*a))*((s^(1/2)*besseli(1 - (a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/a^(1/2) - (besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a*x)))/(-1)^((a + 1)/(2*a)) + (C3*x^((a + 1)/(2*a) - 1)*besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2))*(a + 1))/(2*a) + (C2*x^((a + 1)/(2*a) - 1)*besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2))*(a + 1))/(2*(-1)^((a + 1)/(2*a))*a))/x - s*(C3*x^((a + 1)/(2*a))*besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)) + (C2*x^((a + 1)/(2*a))*besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/(-1)^((a + 1)/(2*a))) 這回我不明白了 |
主管區(qū)長 (職業(yè)作家)
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專家經(jīng)驗: +1059 |

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我在3樓寫的是用matlab算 syms a s y x Y=dsolve('a*D2y-Dy/x-s*y=0','x') 解析解是 Y = C3*x^((a + 1)/(2*a))*besselk(-(a + 1)/(2*a), -(s^(1/2)*x)/a^(1/2)) + (C2*x^((a + 1)/(2*a))*besseli(-(a + 1)/(2*a), (s^(1/2)*x)/a^(1/2)))/(-1)^((a + 1)/(2*a)) 而我在4樓計算的是想通過得到的解析解,反過來計算微分方程右邊是不是等于0 |
主管區(qū)長 (職業(yè)作家)
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專家經(jīng)驗: +1059 |

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