4x^2y'' + y = 0 y'' + y = 0 y_1 = \sqrt{x} y_1 = \sin x
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The given equation is 4x - y" + y = 0. Let's assume the solution has the form y = u(x)v(x), where u(x) is an unknown function and v(x) is the known solution. Substituting this into the equation, we get: 4x - (u''(x)v(x) + 2u'(x)v'(x) + u(x)v''(x)) + u(x)v(x) = Show more…
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