Find a general solution to the following differential equation $x^3y''' - 3x^2y'' + 6xy' - 6y = 3x + 1 + \frac{1}{x}$, $x > 0$
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Let's assume a solution of the form $y = x^m$. Then: $y' = mx^{m-1}$ $y'' = m(m-1)x^{m-2}$ $y''' = m(m-1)(m-2)x^{m-3}$ Substitute these into the homogeneous part of the equation ($x^3y''' - 3x^2y'' + 6xy' - 6y = 0$): $x^3[m(m-1)(m-2)x^{m-3}] - Show more…
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