// Numbas version: finer_feedback_settings {"name": "Ugur's copy of Laplace: Inverse Laplace Completing the Square", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Ugur's copy of Laplace: Inverse Laplace Completing the Square", "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

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Determine the inverse Laplace Transform of the following using completion of the square.

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\\(F(s)=\\dfrac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\)

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", "advice": "

We have

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\\(F(s)=\\frac{\\var{B}s+\\var{C}}{s^2+\\simplify{{b1}*2}s+\\var{c1}}\\).

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By completing the square (on the denominator) we get:

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\\(F(s)=\\frac{\\var{B}s+\\var{C}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}\\).

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Re-grouping the numerator gives:

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\\(F(s)=\\frac{\\var{B}(s+\\var{b1})-\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} = \\frac{\\var{B}(s+\\var{b1})}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} - \\frac{\\simplify{{B}*{b1}-{C}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} = \\frac{\\var{B}(s+\\var{b1})}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}}  - \\frac{\\simplify{{B}*{b1}-{C}}}{\\sqrt{\\simplify{{c1}-{b1}^2}}}\\cdot\\frac{\\sqrt{\\simplify{{c1}-{b1}^2}}}{(s+\\var{b1})^2+\\simplify{{c1}-{b1}^2}} \\)

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Now, the expression above is easy enough to hunt from Laplace Transform table:

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\\(f(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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Write down the inverse Laplace transform

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\\(\\mathscr{L}^{-1}\\{F(s)\\}=\\) [[0]]

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