// Numbas version: finer_feedback_settings {"name": "Differential equation with 3 simple linear factors: I(s)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Differential equation with 3 simple linear factors: I(s)", "tags": [], "metadata": {"description": "
Solve a Differential equation with 3 simple linear factors
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "The solution to the differential equation:
\n\\(\\frac{d^2i}{dt^2}+\\simplify{{a1}+{b1}}\\frac{di}{dt}+\\simplify{{a1}*{b1}}i(t)=\\var{c1}e^{-\\var{d1}t}\\) where \\(i(0)=\\var{i0} \\,\\, and \\,\\, i'(0)=\\var{i1}\\)
\nis given by
\n\\(i(t)=Ae^{-\\var{d1}t}+Be^{-\\var{a1}t}+Ce^{-\\var{b1}t}\\)
", "advice": "\\(\\frac{d^2i}{dt^2}+\\simplify{{a1}+{b1}}\\frac{di}{dt}+\\simplify{{a1}*{b1}}i(t)=\\var{c1}e^{-\\var{d1}t}\\) where \\(i(0)=\\var{i0} \\,\\, and \\,\\, i'(0)=\\var{i1}\\)
\nThe Laplace transform of this is given by:
\n\\(s^2I(s)-\\var{i0}s-\\var{i1}+\\simplify{{a1}+{b1}}(sI(s)-\\var{i0})+\\simplify{{a1}*{b1}}I(s)=\\frac{\\var{c1}}{s+\\var{d1}}\\)
\nGathering only \\(I(s)\\) terms on the left hand side and factoring gives:
\n\\((s^2+\\simplify{{a1}+{b1}}s+\\simplify{{a1}*{b1}})I(s)=\\frac{\\var{c1}}{s+\\var{d1}}+\\var{i0}s+\\simplify{{i1}+({a1}+{b1})*{i0}}\\)
\n\\((s^2+\\simplify{{a1}+{b1}}s+\\simplify{{a1}*{b1}})I(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{b1})*{i0})*(s+{d1})}}{s+\\var{d1}}\\)
\n\\(I(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{b1})*{i0})*(s+{d1})}}{(s+\\var{d1})(s+\\var{a1})(s+\\var{b1})}\\)
\n\\(I(s)=\\frac{A}{s+\\var{d1}}+\\frac{B}{s+\\var{a1}}+\\frac{C}{s+\\var{b1}}\\)
\n\\(\\simplify{{c1}+({i0}s+{i1}+({a1}+{b1})*{i0})*(s+{d1})}=A(s+\\var{a1})(s+\\var{b1})+B(s+\\var{d1})(s+\\var{b1})+C(s+\\var{d1})(s+\\var{a1})\\)
\nLet \\(s=-\\var{d1}\\)
\n\\(\\simplify{{c1}+({i0}*-{d1}+{i1}+({a1}+{b1})*{i0})*(-{d1}+{d1})}=\\simplify{(-{d1}+{a1})(-{d1}+{b1})}A\\)
\n\\(A=\\simplify{({c1}+({i0}*-{d1}+{i1}+({a1}+{b1})*{i0})*(-{d1}+{d1}))/((-{d1}+{a1})(-{d1}+{b1}))}\\)
\nLet \\(s=-\\var{a1}\\)
\n\\(\\simplify{{c1}+({i0}*-{a1}+{i1}+({a1}+{b1})*{i0})*(-{a1}+{d1})}=\\simplify{(-{a1}+{d1})(-{a1}+{b1})}B\\)
\n\\(B=\\simplify{({c1}+({i0}*-{a1}+{i1}+({a1}+{b1})*{i0})*(-{a1}+{d1}))/((-{a1}+{d1})(-{a1}+{b1}))}\\)
\nLet \\(s=-\\var{b1}\\)
\n\\(\\simplify{{c1}+({i0}*-{b1}+{i1}+({a1}+{b1})*{i0})*(-{b1}+{d1})}=\\simplify{(-{b1}+{d1})(-{b1}+{a1})}C\\)
\n\\(C=\\simplify{({c1}+({i0}*-{b1}+{i1}+({a1}+{b1})*{i0})*(-{b1}+{d1}))/((-{b1}+{d1})(-{b1}+{a1}))}\\)
\n\\(i(t)=\\simplify{({c1}+({i0}*-{d1}+{i1}+({a1}+{b1})*{i0})*(-{d1}+{d1}))/((-{d1}+{a1})(-{d1}+{b1}))}e^{-\\var{d1}t}+\\simplify{({c1}+({i0}*-{a1}+{i1}+({a1}+{b1})*{i0})*(-{a1}+{d1}))/((-{a1}+{d1})(-{a1}+{b1}))}e^{-\\var{a1}t}+\\simplify{({c1}+({i0}*-{b1}+{i1}+({a1}+{b1})*{i0})*(-{b1}+{d1}))/((-{b1}+{d1})(-{b1}+{a1}))}e^{-\\var{b1}t}\\)
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