MATH 2214 Midterm: MATH 2214 Virginia Tech 3.7 Exam

• School: Virginia Tech
• Department: Mathematics
• Course: MATH 2214
• Professor: All

Document Summary

Solutions 3. 7: a) y p = 3, y p = 0, y p. 3yp = 6 3(3t 1) = 9t 3: the characteristic equation is r2 2r 3 = 0, which has roots 1 and 3. Hence yc = c1e t + c2e3t: the general solution is. Hence y = 3t 1 + c1e t + c2e3t. y(0) = 1 + c1 + c2 = 1, y (0) = 3 c1 + 3c2 = 3. Hence yc = c1e t + c2. y = 2te t + c1e t + c2. y(0) = c1 + c2 = 2, y (0) = 2 c1 = 2. This leads to c1 = 4, c2 = 6: a) y p = 2 sin t + cos t, y p = 2 cos t sin t, y p. 2y p + 2yp = 5 sin t: the characteristic equation is r2 2r + 2 = 0, which has roots 1 i.