The Given Differential equation is \(x^2 y'' - 3x y' + 4y = \ln(x), \quad x > 0 \) with conditions  y(1) = 6 and y'(1) = 13  then the value of y(e) is ?

1
\(y(e) = 6 e^2 - \frac{1}{7} e^{2} \)
2
\(y(e) = 6 e^2 + \frac{1}{7} e^{-2} \)
3
\(y(e) = 7 e^2 - \frac{1}{6} e^{-2} \)
4
\(y(e) = 7 e^2 + \frac{1}{6} e^{2} \)

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