Lecture 3 Notes

solve y y 0 by assuming yt ert for some constant

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Unformatted text preview: y ￿ + et y = C1 et (et y )￿ = C1 et et y = C1 et + C2 y = C1 + C2 e−t • More common would be that we ﬁnd solutions y(t) = 1 and y(t)= e-t and simply write down −t y = C1 + C2 e Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? • Exponential solutions seem to be common so let’s assume y(t)=ert and see if that gets us anything useful.. Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? • Exponential solutions seem to be common so let’s assume y(t)=ert and see if that gets us anything useful.. • Solve y ￿￿ + y ￿ = 0 by assuming y(t) = ert for some constant r. Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? • Exponential solutions seem to be common so let’s assume y(t)=ert and see if that gets us anything useful.. • Solve y ￿￿ + y ￿ = 0 by assuming y(t) = ert for some constant r. (e ) + (e ) = 0 rt ￿￿ rt ￿ Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? • Exponential solutions seem to be common so let’s assume y(t)=ert and see if that gets us anything useful.. • Solve y ￿￿ + y ￿ = 0 by assuming y(t) = ert for some constant r. (e ) + (e ) = 0 rt ￿￿ rt ￿ r2 ert + rert = 0 Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? • Exponential solutions seem to be common so let’s assume y(t)=ert and see if that gets us anything useful.. • Solve y ￿￿ + y ￿ = 0 by assuming y(t) = ert for some constant r. (e ) + (e ) = 0 rt ￿￿ rt ￿ r2 ert + rert = 0 2 r +r =0 Homog. eq. with constant coeff. (Section 3.1) • So in general how do we ﬁnd the two independent solutions y1 and y2? • Exponential solutions seem to be common so let’s assume y(t)=ert and see if that gets us anything useful.. • Solve y ￿￿ + y ￿ = 0 by assuming y(t) = ert for some constant r. (e ) + (e ) = 0 rt ￿￿ rt ￿ r2 ert + rert...
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This note was uploaded on 02/12/2014 for the course MATH 256 taught by Professor Ericcytrynbaum during the Spring '13 term at The University of British Columbia.

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