※ 引述《dali510313 (乳)》之銘言： : For a differential equation x^2 y" -3xy' +4y = 0, : (a) : use z = lnx to transform such an equation into an equation with constant : coefficients : (b) : find the general solution of (a) in terms of x. : 不好意思,又來打擾各位了。這題讓我想了好久，是假設ｙ的積次因子嗎？ : 看不太懂題意。先謝謝各位了。 原本題目應該是x平方*y'' we don't know the relation between y and x .. dz 1 dx dy but from z = lnx we know ---- = ---- and it can be ---- ---- = y'x = k' dx x dz dx k' means y differential respect to z chain rule 2 dy' d dy dz dz d y dz dy -1 --- = ---- (---- ----) = --- ---- *---- + ---- ---- 2 2 dx dx dz dx dx dz dx dz x from above we know dz/dx = 1/x , so we know y'' = (k'' - k ')/x^2 and y' = k' /x put they into the original equation. x^2 y'' - 3 x y' + 4 y = 0 we get k'' - k' - 3k' + 4y = 0 k'' - 4k' + 4 = 0 get new equation with new constant coefficents. tz slove it , because constant coefficents , use e = y put the solution into the equation. 2 t - 4 t + 4 = 0 t = 2. 2z 2z so the basis of solution is e and ze z and z = ln x so , x = e The general solution is y = c1(x^2) + c2 (lnx*x^2) --

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