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Solusi dari persamaan diferensial
\n$y^{\"}+4y=8sec^{3}(2x)$
\ndengan $y(0)=2$ dan $y'(0)=4$
\nadalah
\nKet:
\ntulis ekspresi fungsinya saja (dalam x)
\nperkalian gunakan * (contoh x*sin(x))
\npangkat dan sinus gunakan tanda kurung (contoh e^(-x), sin(x))
\ninvers trigonometri gunakan arcsin(x), arccos(x), arctan(x)
", "advice": "$y=cos(2x)+2sin(2x)+\\frac{1}{cos(2x)}
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