// Numbas version: finer_feedback_settings {"name": "Marie's copy of Inverse Laplace transform: irreducible quadratic factor", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "rulesets": {}, "statement": "

Having found the partial fraction breakdown:

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\\(Q(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\)

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Write down the inverse Laplace transform

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\\(q(t)=\\) [[0]]

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\\(Q(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\)

\n

\\(Q(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}s+\\var{C}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}\\)

\n

\\(Q(s)=\\frac{\\var{A}}{s+\\var{a1}}+\\frac{\\var{B}(s+\\var{b1})-\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}\\)

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\\(q(t)=\\var{A}e^{\\var{a1}t}+\\var{B}e^{-\\var{b1}t}cos\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)+\\frac{-\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}e^{-\\var{b1}t}sin\\left(\\sqrt{\\simplify{{c1}-{b1}^2}}t\\right)\\)

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