Mathematics, 30.07.2019 21:30 emilyscroggins794
Solve the ivp by variation of parameters: x^2y" - xy' + y = 2x, y(1) = 1, y'(1) = 0
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Solve the ivp by variation of parameters: x^2y" - xy' + y = 2x, y(1) = 1, y'(1) = 0...
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