拉普拉斯反变换 [p*(e^-p)]/1+(p^2) 同时另外一个问题 :Show that,for a > 0,L[x^a] = (a/p)L[x^(a−1)] (1)This was used in class,in conjunction with the result L(1) = 1/p,to show thatL[x^n] = /p^(n+1)where n is a non-negative int
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![拉普拉斯反变换 [p*(e^-p)]/1+(p^2) 同时另外一个问题 :Show that,for a > 0,L[x^a] = (a/p)L[x^(a−1)] (1)This was used in class,in conjunction with the result L(1) = 1/p,to show thatL[x^n] = /p^(n+1)where n is a non-negative int](/uploads/image/z/5100810-42-0.jpg?t=%E6%8B%89%E6%99%AE%E6%8B%89%E6%96%AF%E5%8F%8D%E5%8F%98%E6%8D%A2+%5Bp%2A%28e%5E-p%29%5D%2F1%2B%EF%BC%88p%5E2%EF%BC%89+%E5%90%8C%E6%97%B6%E5%8F%A6%E5%A4%96%E4%B8%80%E4%B8%AA%E9%97%AE%E9%A2%98+%EF%BC%9AShow+that%2Cfor+a+%3E+0%2CL%5Bx%5Ea%5D+%3D+%28a%2Fp%29L%5Bx%5E%28a%26%238722%3B1%29%5D+%281%29This+was+used+in+class%2Cin+conjunction+with+the+result+L%281%29+%3D+1%2Fp%2Cto+show+thatL%5Bx%5En%5D+%3D+%2Fp%5E%28n%2B1%29where+n+is+a+non-negative+int)
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