MATH 2214 Midterm: MATH 2214 Virginia Tech 3.4 Exam

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

Document Summary

Solutions 3. 4: a) the characteristic equation is. 6 + 1 = 0, leading to = 1/3. The general solution is y = c1et/3 + c2tet/3. b) y(3) = ec1 + 3ec2 = 2, y (3) = e. 3 c1 + 2ec2 = 5/3, hence c1 = 1/e, c2 = 1/e: limt y(t) = 0, limt y(t) = , a) the characteristic equation is. 4 + 4 = 0, leading to = 2. 5 + 6. 25 = 0, leading to = 5/2. The general solution is y = c1e5t/2 + c2te5t/2. b) y( 2) = e 5c1 2e 5c2 = 0, y ( 2) = 2 e 5c1 4e 5c2 = 1, hence c1 = 2e5, c2 = e5: limt y(t) = 0, limt y(t) = + .