Question 3. [8 points] Find the general solution of the following ODE: $x^2y'' - 3xy' + 3y = 2x^4 \ln(x)$, $x > 0$.
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The given equation is xy - 3xy + 3y = 2xln(x) > 0. Simplifying the equation, we get -2xy + 3y = 2xln(x) > 0. Show more…
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