Discussion
Let \(y(x)\) be the solution of the differential equation
\[ x^{2}y'' + 7xy' + 9y = x^{-3}\log_{e} x, \quad x \gt 0, \]
satisfying \(y(1) = 0\) and \(y'(1) = 0\). Then, the value of \(y(e)\) is equal to
Answer: B
The value is \(\dfrac{1}{6}e^{-3}\).
This is a Cauchy-Euler equation. With \(x = e^{t}\) it becomes
\[ \begin{aligned} y'' + 6y' + 9y &= te^{-3t} \\ (D + 3)^{2}y &= te^{-3t} \end{aligned} \]
The root -3 is double, so try \(y_{p} = At^{3}e^{-3t}\): \((D + 3)^{2}(t^{3}e^{-3t}) = 6te^{-3t}\), giving A = 1/6. Thus
\[ y = (c_{1} + c_{2}t)e^{-3t} + \tfrac{1}{6}t^{3}e^{-3t} \]
x = 1 is t = 0, and the conditions give \(c_{1} = 0\) and \(c_{2} = 0\). At x = e, t = 1, so \(y(e) = \tfrac{1}{6}e^{-3}\). Missing the factor 6 from the double root leads to the other fractions.
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