Unformatted text preview: • What is the solution to the associated homogeneous equation?
(A) yh (t) = C1 e + C2 e (B) yh (t) = C1 cos(2t) + C2 sin(2t) 2t (C) yh (t) −2t = C1 e + C2 te
2t 2t (D) yh (t) = C1 e + C2 e (E) yh (t) = C1 cos(2t) + C2 sin(2t) + e 2t −2t +e t
t Method of undetermined coefﬁcients (3.5)
− 4y = e2t. • Example 6. Find the general solution to the equation y • What is the solution to the associated homogeneous equation?
(A) yh (t) = C1 e + C2 e (B) yh (t) = C1 cos(2t) + C2 sin(2t) 2t −2t am
ex t 2t
Cs
(C) yh (t) = C1 e + a2 te
l
hte −2t t
t2
as 1 e + C2 e + e
(D) yh (t) = C
e
am
S
2t (E) le
p yh (t) = C1 cos(2t) + C2 sin(2t) + e t Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y
• What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate (D) yp (t) = Ae (E) yp (t) = Ate 2t
−2t
2t t
t − 4y = e2t. Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y − 4y = e2t. • What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate (D) yp (t) = Ae (E) yp (t) = Ate 2t
−2t
2t t
t (Ae2t ) − 4Ae2t = 0 ! Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y − 4y = e2t. • What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate 2t
−2t
2t (Ae2t ) − 4Ae2t = 0 !
• Simpler example in which
the RHS is a solution to the
homogeneous problem. y −y =e
(D) yp (t) = Ae (E) yp (t) = Ate t
t t Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y − 4y = e2t. • What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate 2t
−2t
2t (Ae2t ) − 4Ae2t = 0 !
• Simpler example in which
the RHS is a solution to the
homogeneous problem. y −y =e
(D) yp (t) = Ae (E) yp (t) = Ate t
t t e−t y − e−t y = 1 Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y − 4y = e2t. • What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate 2t
−2t
2t (Ae2t ) − 4Ae2t = 0 !
• Simpler example in which
the RHS is a solution to the
homogeneous problem. y −y =e
(D) yp (t) = Ae (E) yp (t) = Ate t
t t e−t y − e−t y = 1
y = tet + Cet Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y − 4y = e2t. • What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate 2t
−2t
2t (Ae2t ) − 4Ae2t = 0 !
• Simpler example in which
the RHS is a solution to the
homogeneous problem. y −y =e
(D) yp (t) = Ae (E) yp (t) = Ate t
t t e−t y − e−t y = 1
y = tet + Cet Method of undetermined coefﬁcients (3.5)
• Example 6. Find the general solution to the equation y − 4y = e2t. • What is the form of the particular solution?
(A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate 2t
−2t
2t (Ae2t ) − 4Ae2t = 0 !
• Simpler example in which
the RHS is a solution to the
homogeneous problem. y −y =e
(D) yp (t) = Ae (E) yp (t) = Ate t
t t e−t y − e−t y = 1
y = tet + Cet 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 )? 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)...
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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 UBC.
 Spring '13
 EricCytrynbaum
 Differential Equations, Geometry, Equations

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