Lecture 1 Notes

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online