Lecture 6 Notes

Method of undetermined coefcients 35 example 6 find

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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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