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The solution to the differential equation:

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     \\(\\frac{d^2x}{dt^2}+\\simplify{{a1}+{a1}}\\frac{dx}{dt}+\\simplify{{a1}*{a1}}x(t)=\\var{c1}e^{-\\var{d1}t}+\\var{b}\\delta(t-\\var{f})\\)  where \\(x(0)=\\var{i0} \\,\\, and \\,\\,  x'(0)=\\var{i1}\\)

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is given by

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     \\(x(t)=Ae^{-\\var{d1}t}+Be^{-\\var{a1}t}+Cte^{-\\var{a1}t}+Du(t-\\var{f})(t-\\var{f})e^{-\\var{a1}(t-\\var{f})}\\)

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\\(\\frac{d^2x}{dt^2}+\\simplify{{a1}+{a1}}\\frac{dx}{dt}+\\simplify{{a1}*{a1}}x(t)=\\var{c1}e^{-\\var{d1}t}+\\var{b}\\delta(t-\\var{f})\\)  where \\(x(0)=\\var{i0} \\,\\, and \\,\\,  x'(0)=\\var{i1}\\)

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The Laplace transform of this is given by:

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\\(s^2X(s)-\\var{i0}s-\\var{i1}+\\simplify{{a1}+{a1}}(sX(s)-\\var{i0})+\\simplify{{a1}*{a1}}X(s)=\\frac{\\var{c1}}{s+\\var{d1}}+\\var{b}e^{-\\var{f}s}\\)

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Gathering only \\(X(s)\\) terms on the left hand side and factoring gives:

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\\((s^2+\\simplify{{a1}+{a1}}s+\\simplify{{a1}*{a1}})X(s)=\\frac{\\var{c1}}{s+\\var{d1}}+\\var{i0}s+\\simplify{{i1}+({a1}+{a1})*{i0}}+\\var{b}e^{-\\var{f}s}\\)

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\\((s^2+\\simplify{{a1}+{a1}}s+\\simplify{{a1}*{a1}})X(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{a1})*{i0})*(s+{d1})}}{s+\\var{d1}}+\\var{b}e^{-\\var{f}s}\\)

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\\(X(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{a1})*{i0})*(s+{d1})}}{(s+\\var{d1})(s+\\var{a1})^2}+\\frac{\\var{b}}{(s+\\var{a1})^2}e^{-\\var{f}s}\\)

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Solving the first fraction gives:

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\\(X(s)=\\frac{A}{s+\\var{d1}}+\\frac{B}{s+\\var{a1}}+\\frac{C}{(s+\\var{a1})^2}\\)

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\\(\\simplify{{c1}+({i0}s+{i1}+({a1}+{a1})*{i0})*(s+{d1})}=A(s+\\var{a1})(s+\\var{a1})+B(s+\\var{d1})(s+\\var{a1})+C(s+\\var{d1})\\)

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Let \\(s=-\\var{d1}\\)

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\\(\\simplify{{c1}+({i0}*-{d1}+{i1}+({a1}+{a1})*{i0})*(-{d1}+{d1})}=\\simplify{(-{d1}+{a1})(-{d1}+{a1})}A\\)

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\\(A=\\simplify{({c1})/((-{d1}+{a1})(-{d1}+{a1}))}\\)

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Let \\(s=-\\var{a1}\\)

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\\(\\simplify{{c1}+({i0}*-{a1}+{i1}+({a1}+{a1})*{i0})*(-{a1}+{d1})}=\\simplify{(-{a1}+{d1})}C\\)

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\\(C=\\simplify{({c1}+({i0}*-{a1}+{i1}+({a1}+{a1})*{i0})*(-{a1}+{d1}))/((-{a1}+{d1}))}\\)

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Compare the coefficients of \\(s^2\\)   

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\\(\\var{i0}=A+B\\)

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\\(B=\\simplify{{i0}-(({c1})/((-{d1}+{a1})(-{d1}+{a1})))}\\)

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We can find the inverse Laplace transform of the second fraction without breaking it down: 

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\\(\\frac{\\var{b}}{(s+\\var{a1})^2}e^{-\\var{f}s}\\) changes to \\(\\var{b}u(t-\\var{f})(t-\\var{f})e^{-\\var{a1}(t-\\var{f})}\\)

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\\(D=\\var{b}\\)

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Enter the value for \\(A\\) correct to three decimal places.

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Enter the value for \\(B\\) correct to three decimal places.

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Enter the value for \\(C\\) correct to three decimal places.

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Enter the value for \\(D\\).

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Solve a Differential equation with a repeated linear factor & a delta function

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