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Diberikan persamaan diferensial $\\begin{cases} y''-2y'+y=0, & x<0 \\\\ y''+y'-2y=0, & x\\geq 0 \\end{cases}$.

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Jika diketahui $y(0)=1$, $y'(0)=\\var{3*k+1}$, dan $y$ serta $y'$ adalah fungsi-fungsi yang kontinu untuk setiap $x\\in \\mathbb{R}$, maka $y(-1)=\\dfrac{k}{e}$ dengan $k=\\ldots$.

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