// Numbas version: exam_results_page_options {"name": "Differential Equations: Second Order Repeated Roots", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Differential Equations: Second Order Repeated Roots", "tags": [], "metadata": {"description": "
Solve: $\\displaystyle \\frac{d^2y}{dx^2}+2a\\frac{dy}{dx}+a^2y=0,\\;y(0)=c$ and $y(1)=d$. (Equal roots example).
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the solution of the differential equation:
\n
\\[\\dfrac{d^2y}{dx^2}+\\var{2*a}\\dfrac{dy}{dx}+\\var{a^2}y=0\\]
which satisfies $y(0)=\\var{c}$ and $y(1)=\\var{d}$.
The auxillary equation is $\\simplify[std]{lambda^2+{2*a}lambda+{a^2}}=0$.
\nOn solving this equation we get $\\lambda=\\var{-a}$ twice.
\nHence the general solution is:
\\[y = \\simplify[std]{A*e^({-a}x)+B*x*e^({-a}x)}\\]
The boundary conditions give:
$y(0)=\\var{c} \\Rightarrow A=\\var{c}$
\n$y(1)=\\var{d} \\Rightarrow \\simplify{Ae^{-a}+Be^{-a}={d}}\\Rightarrow A+B = \\simplify{{d}e^{a}}$
\nSo $B=\\simplify{{d}e^{a}-{c}}=\\var{f1}$ to 3 decimal places.
\nHence the solution is:
\\[y=\\simplify{(({c} * Exp(({( - a)} * x))) + ({f1} * x * Exp(({( - a)} * x))))}\\]
Solution is:
\n$y=\\;\\;$[[0]]
\nInput all numbers correct to 3 decimal places.
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