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Diiberikan persamaan diferensial \$x''-\\var{2*k}x'+\\var{k^2}x=(2t+\\var{k})e^{\\var{k}t}\\sec(t^2+\\var{k}t)\$ dengan syarat awal \$x(0)=1\$ dan \$x'(0)=2\$.

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Jika solusi dari persamaan diferensial tersebut adalah \$x(t)\$, maka \$x''(0)=\\ldots\$.

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