Lecture 1 Notes

1 a y t 2 sint t b y t cost c c cost c

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Unformatted text preview: on 2.1) ￿ d￿2 t y (t) = dt dy (A) 2t dt (B) t 2 dy dt (C) 2ty (D) t 2 dy dt + 2ty Method of integrating factors (Section 2.1) • Given that ￿ d￿2 2 dy t y (t) = t + 2ty dt dt • if you’re given the equation dt + 2ty = 0 ￿ d￿2 t y (t) = 0 dt • you can rewrite is as • so the solution is t 2 dy t y (t) = C 2 or equivalently arbitrary constant that appeared at an integration step C y (t) = 2 t . Method of integrating factors (Section 2.1) • Solve the equation t 2 dy dt + 2ty (t) = sin(t) 1 (A) y (t) = − 2 sin(t) t (B) y (t) = − cos(t) + C C − cos(t) (C) y (t) = t2 (D) y (t) = sin(t) + C 1 (E) y (t) = − 2 cos(t) t (not brute force checking). Method of integrating factors (Section 2.1) • Solve the equation t 2 dy dt + 2ty (t) = sin(t) (not brute force checking). 1 (A) y (t) = − 2 sin(t) t (B) y (t) = − cos(t) + C C − cos(t) (C) y (t) = t2 (D) y (t) = sin(t) + C 1 (E) y (t) = − 2 cos(t) t general solution (although that’s not obvious) a particular solution Initial conditions (IC) and initial value problems (IVP) Initial conditions (IC) and initial value problems (IVP) • An initial condition is an added constraint on a solution. Initial conditions (IC) and initial val...
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