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Solve a Differential equation with a  simple irreducible quadratic factor

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\$$\\frac{d^2q}{dt^2}+\\simplify{{a1}^2}q(t)=\\var{c1}e^{-\\var{d1}t}\$$  where \$$q(0)=\\var{i0} \\,\\, and \\,\\, q'(0)=\\var{i1}\$$

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The Laplace transform of this is given by:

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\$$s^2Q(s)-\\var{i0}s-\\var{i1}+\\simplify{{a1}^2}Q(s)=\\frac{\\var{c1}}{s+\\var{d1}}\$$

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

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\$$(s^2+\\simplify{{a1}^2})Q(s)=\\frac{\\var{c1}}{s+\\var{d1}}+\\var{i0}s+\\var{i1}\$$

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\$$(s^2+\\simplify{{a1}*{a1}})Q(s)=\\frac{\\simplify{{c1}+({i0}s+{i1})*(s+{d1})}}{s+\\var{d1}}\$$

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\$$Q(s)=\\frac{\\simplify{{c1}+({i0}s+{i1})*(s+{d1})}}{(s+\\var{d1})(s^2+\\var{a1}^2)}\$$

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\$$Q(s)=\\frac{A}{s+\\var{d1}}+\\frac{Bs+c}{s^2+\\var{a1}^2}\$$

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\$$\\simplify{{c1}+({i0}s+{i1})*(s+{d1})}=A(s^2+\\var{a1}^2)+Bs(s+\\var{d1})+c(s+\\var{d1})\$$

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Let \$$s=-\\var{d1}\$$

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\$$\\var{c1}=\\simplify{{d1}^2+{a1}^2}A\$$

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\$$A=\\simplify{({c1})/(({d1}^2+{a1}^2))}\$$

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Let \$$s=0\$$

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\$$\\simplify{{c1}+{i1}*{d1}}=\\simplify{{a1}^2}A+{\\var{d1}}c\$$

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\$$\\simplify{{c1}+{i1}*{d1}}=\\simplify{{a1}^2*{c1}/({d1}^2+{a1}^2)}+{\\var{d1}}c\$$

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\$$\\var{d1}c=\\simplify{{c1}+{i1}*{d1}}-\\simplify{{a1}^2*{c1}/({d1}^2+{a1}^2)}\$$

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\$$c=\\frac{\\simplify{({c1}+{i1}*{d1})-({a1}^2*{c1})/({d1}^2+{a1}^2)}}{\\var{d1}}\$$

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But the coefficient of \$$sin(\\var{a1}t)\$$ is given by \$$C=\\frac{c}{\\var{a1}}\$$

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\$$C=\\frac{\\simplify{{c1}+{i1}*{d1}-{a1}^2*{c1}/({d1}^2+{a1}^2)}}{\\simplify{{a1}*{d1}}}\$$

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\$$C=\\simplify{(({c1}+{i1}*{d1})*({d1}^2+{a1}^2)-{a1}^2*{c1})/(({d1}^2+{a1}^2)*({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=\\var{i0}-\\simplify{({c1})/(({d1}^2+{a1}^2))}=\\simplify{(({d1}^2+{a1}^2)*{i0}-{c1})/({d1}^2+{a1}^2)}\$$

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", "tags": [], "statement": "

The solution to the differential equation:

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\$$\\frac{d^2q}{dt^2}+\\simplify{{a1}^2}q(t)=\\var{c1}e^{-\\var{d1}t}\$$  where \$$q(0)=\\var{i0} \\,\\, and \\,\\, q'(0)=\\var{i1}\$$

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is given by

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\$$q(t)=Ae^{-\\var{d1}t}+Bcos(\\var{a1}t)+Csin(\\var{a1}t)\$$

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