Lecture 6 Notes

5 example 4 dene the operator general solution to ly

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Unformatted text preview: r general solution to L[y ] = y + 2y − 3y. ￿￿ L[y ] = e 2t . That is, • Summarizing: ￿ Find the y ￿￿ + 2y ￿ − 3y = e2t . • We know that, for any C1 and C2, L[C1 et + C2 e−3t ] = 0 L[C1 e + C2 e t −3t + Ae ] = 0 + 5Ae 2t 2t Method of undetermined coefficients (3.5) • Example 4. Define the operator general solution to L[y ] = y + 2y − 3y. ￿￿ L[y ] = e 2t . That is, • Summarizing: ￿ Find the y ￿￿ + 2y ￿ − 3y = e2t . • We know that, for any C1 and C2, L[C1 et + C2 e−3t ] = 0 • We also know that L[Ae2t ] = 5Ae2t L[C1 e + C2 e t −3t + Ae ] = 0 + 5Ae 2t 2t Method of undetermined coefficients (3.5) • Example 4. Define the operator general solution to L[y ] = y + 2y − 3y. ￿￿ L[y ] = e 2t . That is, • Summarizing: ￿ Find the y ￿￿ + 2y ￿ − 3y = e2t . • We know that, for any C1 and C2, L[C1 et + C2 e−3t ] = 0 • We also know that L[Ae2t ] = 5Ae2t • Finally, by linearity, we know that L[C1 e + C2 e t −3t + Ae ] = 0 + 5Ae 2t 2t Method of undetermined coefficients (3.5) • Example 4. Define the operator general solution to L[y ] = y + 2y − 3y. ￿￿ L[y ] = e 2t . That is, • Summarizing: ￿ Find the y ￿￿ + 2y ￿ − 3y = e2t . • We know that, for any C1 and C2, L[C1 et + C2 e−3t ] = 0 • We also know that L[Ae2t ] = 5Ae2t • Finally, by linearity, we know that L[C1 e + C2 e t −3t + Ae ] = 0 + 5Ae 2t 2t • So what’s left to do to find our general solution? Pick A =? Method of undetermined coefficients (3.5) • Example 4. Define the operator general solution to L[y ] = y + 2y − 3y. ￿￿ L[y ] = e 2t . That is, • Summarizing: ￿ Find the y ￿￿ + 2y ￿ − 3y = e2t . • We know that, for any C1 and C2, L[C1 et + C2 e−3t ] = 0 • We also know that L[Ae2t ] = 5Ae2t • Finally, by linearity, we know that L[C1 e + C2 e t −3t + Ae ] = 0 + 5Ae 2t 2t • So what’s left to do to find our general solution? Pick A = 1/5. Method of undetermined coefficients (3.5) • Example 5. Find the general solution to the equation y ￿￿ − 4y = et. • 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) • Example 5. Find the general solution to the equation y ￿￿ − 4y = et. • 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) • Example 5. Find the general solution to the equation • What is the form of the particular solution? (A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate−2t (D) yp (t) = Ae (E) yp (t) = Ate 2t −2t t t y ￿￿ − 4y = et. Method of undetermined coefficients (3.5) • Example 5. Find the general solution to the equation • What is the form of the particular solution? (A) yp (t) = Ae (B) yp (t) = Ae (C) yp (t) = Ate−2t (D) yp (t) = Ae (E) yp (t) = Ate 2t −2t t t y ￿￿ − 4y = et. Method of undetermined coefficients (3.5) • Example 5. Find the general solution to the equation y ￿￿ − 4y = et. • What is the value of A that gives the particular solution (Ae t (A) A = 1 (B) A = 3 (C) A = -3 (D) A = 1/3 (E) A = -1/3 )? Method of undetermined coefficients (3.5) • Example 5. Find the general solution to the equation y ￿￿ − 4y = et. • What is the value of A that gives the particular solution (Ae t (A) A = 1 (B) A = 3 (C) A = -3 (D) A = 1/3 (E) A = -1/3 )? Method of undetermined coefficients (3.5) − 4y = e2t. • Example 6. Find the general solution to the equation y ...
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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.

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