This preview shows page 1. Sign up to view the full content.
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)= et 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...
View
Full
Document
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.
 Spring '13
 EricCytrynbaum
 Differential Equations, Equations

Click to edit the document details