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\\(\\frac{d^2x}{dt^2}+\\var{a}\\frac{dx}{dt}+\\var{b}x(t)=\\var{c}e^{-\\var{d}t}\\)    where   \\(x(0)=\\var{f}\\) and  \\(x'(0)=\\var{g}\\)

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\\(s^2X(s)-sx(0)-x'(0)+\\var{a}(s(X(s)-x(0))+\\var{b}X(s)=\\frac{\\var{c}}{s+\\var{d}}\\)

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\\(s^2X(s)-\\var{f}s-\\var{g}+\\var{a}sX(s)-\\var{a}*\\var{f}+\\var{b}X(s)=\\frac{\\var{c}}{s+\\var{d}}\\)

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\\(s^2X(s)+\\var{a}sX(s)+\\var{b}X(s)=\\frac{\\var{c}}{s+\\var{d}}+\\var{f}s+\\simplify{{g}+{a}*{f}}\\)

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\\((s^2+\\var{a}s+\\var{b})X(s)=\\frac{\\var{c}+(\\var{f}s+\\simplify{{g}+{a}*{f}})(s+\\var{d})}{s+\\var{d}}\\)

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\\(X(s)=\\frac{\\simplify{{f}s^2+({a}*{f}+{g}+{d}*{f})s+(({g}+{f}*{a})*{d}+{c})}}{(s+\\var{d})(s^2+\\var{a}s+\\var{b})}\\)

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Given the differential equation:

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\\(\\dfrac{d^2x}{dt^2}+\\var{a}\\dfrac{dx}{dt}+\\var{b}x=\\var{c}e^{-\\var{d}t}\\)    where   \\(x(0)=\\var{f}\\) and  \\(x'(0)=\\var{g}\\)

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  1. Take the Laplace transform of the equation,
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  3. Substitute in the conditions,
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  5. Rearrange for \\(\\mathscr{L}(x)\\) and then express the right-hand side as a single fraction.
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  7. Enter this fraction in the box below.
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\\(\\mathscr{L}(x)=\\) [[0]]

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