// Numbas version: exam_results_page_options {"name": "Differential equation with a repeated linear factor & a delta function: I(s)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"condition": "", "maxRuns": "222"}, "name": "Differential equation with a repeated linear factor & a delta function: I(s)", "variable_groups": [], "metadata": {"description": "

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

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Enter the value for \\(A\\) as an exact fraction.

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Enter the value for \\(B\\) as an exact fraction.

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Enter the value for \\(C\\) as an exact fraction.

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

", "mustBeReduced": false, "minValue": "b", "correctAnswerStyle": "plain", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "mustBeReducedPC": 0, "variableReplacements": []}], "variables": {"i1": {"description": "", "templateType": "anything", "definition": "random(1..10) ", "name": "i1", "group": "Ungrouped variables"}, "b": {"description": "", "templateType": "randrange", "definition": "random(2..12#1)", "name": "b", "group": "Ungrouped variables"}, "a1": {"description": "", "templateType": "anything", "definition": "random(1..5)", "name": "a1", "group": "Ungrouped variables"}, "d1": {"description": "", "templateType": "anything", "definition": "random(12..16)", "name": "d1", "group": "Ungrouped variables"}, "i0": {"description": "", "templateType": "anything", "definition": "random(1..10)", "name": "i0", "group": "Ungrouped variables"}, "f": {"description": "", "templateType": "randrange", "definition": "random(1..9#1)", "name": "f", "group": "Ungrouped variables"}, "c1": {"description": "", "templateType": "anything", "definition": "random(1..10)", "name": "c1", "group": "Ungrouped variables"}}, "tags": [], "statement": "

The solution to the differential equation:

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

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is given by

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     \\(i(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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.

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

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The Laplace transform of this is given by:

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

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

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\\((s^2+\\simplify{{a1}+{a1}}s+\\simplify{{a1}*{a1}})I(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}})I(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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\\(I(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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\\(I(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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\\(B=\\simplify{(({i0}*(-{d1}+{a1})*(-{d1}+{a1})-{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}\\)

", "functions": {}, "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}