x^2y'' - xy' + 2y = 0; y_1 = x sin(ln x) Answer: y_2 = x cos(ln x)
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The given differential equation is \(x^2y'' - xy' + 2y = 0\), where \(y''\) denotes the second derivative of \(y\) with respect to \(x\), and \(y'\) denotes the first derivative of \(y\) with respect to \(x\). We are also given a solution \(y_1 = x \sin(\ln(x))\) Show more…
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