// Numbas version: exam_results_page_options
{"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": "<p>Solve a&nbsp;Differential equation&nbsp;with 3 simple linear factors</p>", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "<p>The solution to the differential equation:</p>\n<p>&nbsp; &nbsp; &nbsp;\\(\\frac{d^2i}{dt^2}+\\simplify{{a1}+{b1}}\\frac{di}{dt}+\\simplify{{a1}*{b1}}i(t)=\\var{c1}e^{-\\var{d1}t}\\) &nbsp;where \\(i(0)=\\var{i0} \\,\\, and \\,\\, &nbsp;i'(0)=\\var{i1}\\)</p>\n<p>is given by</p>\n<p>&nbsp; &nbsp; &nbsp;\\(i(t)=Ae^{-\\var{d1}t}+Be^{-\\var{a1}t}+Ce^{-\\var{b1}t}\\)</p>", "advice": "<p>\\(\\frac{d^2i}{dt^2}+\\simplify{{a1}+{b1}}\\frac{di}{dt}+\\simplify{{a1}*{b1}}i(t)=\\var{c1}e^{-\\var{d1}t}\\) &nbsp;where \\(i(0)=\\var{i0} \\,\\, and \\,\\, &nbsp;i'(0)=\\var{i1}\\)</p>\n<p>The Laplace transform of this is given by:</p>\n<p>\\(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}}\\)</p>\n<p>Gathering only \\(I(s)\\) terms on the left hand side and factoring gives:</p>\n<p>\\((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}}\\)</p>\n<p>\\((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}}\\)</p>\n<p>\\(I(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{b1})*{i0})*(s+{d1})}}{(s+\\var{d1})(s+\\var{a1})(s+\\var{b1})}\\)</p>\n<p>\\(I(s)=\\frac{A}{s+\\var{d1}}+\\frac{B}{s+\\var{a1}}+\\frac{C}{s+\\var{b1}}\\)</p>\n<p>\\(\\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})\\)</p>\n<p>Let \\(s=-\\var{d1}\\)</p>\n<p style=\"padding-left: 30px;\">\\(\\simplify{{c1}+({i0}*-{d1}+{i1}+({a1}+{b1})*{i0})*(-{d1}+{d1})}=\\simplify{(-{d1}+{a1})(-{d1}+{b1})}A\\)</p>\n<p style=\"padding-left: 30px;\">\\(A=\\simplify{({c1}+({i0}*-{d1}+{i1}+({a1}+{b1})*{i0})*(-{d1}+{d1}))/((-{d1}+{a1})(-{d1}+{b1}))}\\)</p>\n<p>Let \\(s=-\\var{a1}\\)</p>\n<p style=\"padding-left: 30px;\">\\(\\simplify{{c1}+({i0}*-{a1}+{i1}+({a1}+{b1})*{i0})*(-{a1}+{d1})}=\\simplify{(-{a1}+{d1})(-{a1}+{b1})}B\\)</p>\n<p style=\"padding-left: 30px;\">\\(B=\\simplify{({c1}+({i0}*-{a1}+{i1}+({a1}+{b1})*{i0})*(-{a1}+{d1}))/((-{a1}+{d1})(-{a1}+{b1}))}\\)</p>\n<p>Let \\(s=-\\var{b1}\\)</p>\n<p style=\"padding-left: 30px;\">\\(\\simplify{{c1}+({i0}*-{b1}+{i1}+({a1}+{b1})*{i0})*(-{b1}+{d1})}=\\simplify{(-{b1}+{d1})(-{b1}+{a1})}C\\)</p>\n<p style=\"padding-left: 30px;\">\\(C=\\simplify{({c1}+({i0}*-{b1}+{i1}+({a1}+{b1})*{i0})*(-{b1}+{d1}))/((-{b1}+{d1})(-{b1}+{a1}))}\\)</p>\n<p>\\(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}\\)</p>", "rulesets": {}, 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