Lecture 6 Notes

# A yp t a cos2t b yp t a sin2t c yp t

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Unformatted text preview: A = -4 (D) A = 1/4 (E) A = -1/4 ￿ ￿ 2t ￿￿ Ate ￿ − 4 Ate 2t ￿ = )? Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to the equation y ￿￿ − 4y = e2t. • What is the value of A that gives the particular solution (Ate 2t (A) A = 1 (B) A = 4 (C) A = -4 (D) A = 1/4 (E) A = -1/4 ￿ ￿ 2t ￿￿ Ate ￿ − 4 Ate 2t ￿ = 4Ae2t )? Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to the equation y ￿￿ − 4y = e2t. • What is the value of A that gives the particular solution (Ate 2t (A) A = 1 (B) A = 4 (C) A = -4 (D) A = 1/4 (E) A = -1/4 ￿ ￿ 2t ￿￿ Ate ￿ − 4 Ate 2t ￿ = 4Ae2t )? Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y − 4y = cos(2t) . ￿￿ • What is the form of the particular solution? (A) yp (t) = A cos(2t) (B) yp (t) = A sin(2t) (C) yp (t) = A cos(2t) + B sin(2t) (D) yp (t) = t(A cos(2t) + B sin(2t)) (E) yp (t) = e (A cos(2t) + B sin(2t)) 2t Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y − 4y = cos(2t) . ￿￿ • What is the form of the particular solution? (A) yp (t) = A cos(2t) (B) yp (t) = A sin(2t) (C) yp (t) = A cos(2t) + B sin(2t) (D) yp (t) = t(A cos(2t) + B sin(2t)) (E) yp (t) = e (A cos(2t) + B sin(2t)) 2t Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y − 4y = cos(2t) . ￿￿ • What is the form of the particular solution? (A) yp (t) = A cos(2t) (B) yp (t) = A sin(2t) (C) yp (t) = A cos(2t) + B sin(2t) (D) yp (t) = t(A cos(2t) + B sin(2t)) (E) yp (t) = e (A cos(2t) + B sin(2t)) 2t What small change to the DE makes (D) correct? Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y + y − 4y = cos(2t). ￿￿ ￿ • What is the form of the particular solution? (A) yp (t) = A cos(2t) (B) yp (t) = A sin(2t) (C) yp (t) = A cos(2t) + B sin(2t) (D) yp (t) = t(A cos(2t) + B sin(2t)) (E) yp (t) = e (A cos(2t) + B sin(2t)) 2t Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y + y − 4y = cos(2t). ￿￿ ￿ • What is the form of the particular solution? (A) yp (t) = A cos(2t) (B) yp (t) = A sin(2t) (C) yp (t) = A cos(2t) + B sin(2t) (D) yp (t) = t(A cos(2t) + B sin(2t)) (E) yp (t) = e (A cos(2t) + B sin(2t)) 2t Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y ￿￿ − 4y = t3 . • What is the form of the particular solution? (A) yp (t) = At (B) yp (t) = At + Bt + Ct (C) yp (t) = At3 + Bt2 + Ct + D (D) yp (t) = At3 + Be2t + Ce−2t (E) yp (t) = At3 + Bt2 + Ct + D + Ee2t + F e−2t 3 3 2 Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y ￿￿ − 4y = t3 . • What is the form of the particular solution? (A) yp (t) = At (B) yp (t) = At + Bt + Ct (C) yp (t) = At3 + Bt2 + Ct + D (D) yp (t) = At3 + Be2t + Ce−2t (E) yp (t) = At3 + Bt2 + Ct + D + Ee2t + F e−2t 3 3 2 Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y + 2y = e + t ￿￿ ￿ 2t 3 . • What is the form of the particular solution? (A) yp (t) = Ae2t + Bt3 + Ct2 + Dt (B) yp (t) = Ae2t + Bt3 + Ct2 + Dt + E (C) yp (t) = Ae2t + (Bt4 + Ct3 + Dt2 + Et) (D) yp (t) = Ae2t + Be−2t + Ct3 + Dt2 + Et + F (E) yp (t) = Ae + Bte + Ct + Dt + Et + F 2t 2t 3 2 Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y + 2y = e + t ￿￿ ￿ 2t 3 . • What is the form of the particular solution? (A) yp (t) = Ae2t + Bt3 + Ct2 + Dt (B) yp (t) = Ae2t + Bt3 + Ct2 + Dt + E (C) yp (t) = Ae2t + (Bt4 + Ct3 + Dt2 + Et) (D) yp (t) = Ae2t + Be−2t + Ct3 + Dt2 + Et + F (E) yp (t) = Ae + Bte + Ct + Dt + Et + F 2t 2t 3 2 Method of undetermined coefﬁcients (3.5) • Example 6. Find the general solution to y + 2y = e + t ￿￿ ￿ 2t 3 . • What is the form of the particular solution...
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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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