Quiz 1 Solution
Math 307 G. January 23, 2014
Problem 1. Find the general solution to the equation
y = (y 2)4 e2t .
Solution. Separation of variables: assume y = 2. Then
dy
dy
= (y 2)4 e2t
= e2t dt
4
dt
(y 2)
But we know that
z a dz =
dy
=
(y 2)4
e2t dt.
Quiz 5 Solution
Math 307 G. February 20, 2015
Problem 1. Find the general solution to the equation
y + 4y + 5y = (2 t)et cos(2t) + (2t2 + 3)e2t sin(t) t2 e3t + 5.
Do not nd undetermined coecients, just leave them as A, B, . . .
Solution. First, let us sol
Quiz 4 Solution
Math 307 G. February 18, 2015
Problem 1. Find the general solution to the equation
y + 2y + y = (5t3 + 2t2 )et + 3t3 e2t t4 .
Do not nd undetermined coecients, just leave them as A, B, . . .
Solution. First, let us solve the homogeneous eq
Quiz 4 Solution
Math 307 J. February 18, 2015
Problem 1. Find the general solution to the equation
2y 8y = (t3 + 5t)e2t + te2t 2t2 + 5et .
Do not nd undetermined coecients, just leave them as A, B, . . .
Solution. First, let us solve the homogeneous equat
Quiz 1 Solution
Math 307 J. January 23, 2014
Problem 1. Find the general solution to the equation
y = (1 + y) sin t.
Solution. Method 1. We can write this as a linear equation:
y = sin t + y sin t.
First, solve the corresponding homogeneous equation:
y =
Quiz 3
Math 307 J. February 6, 2015
Problem 1. Find the general solution to the equation
y + 4y + 8y = 0.
Solution. Characteristic equation: 2 + 4 + 8 = 0. Its roots:
4 42 4 8
4 16
=
=
= 2 2i.
2
2
Therefore, the general solution is
y = C1 e2t cos(2t) + C2
Quiz 3 Solution
Math 307 G. February 6, 2015
Problem 1. Find the general solution to the equation
y + 4y + 4y = 0.
Solution. Characteristic equation: 2 + 4 + 4 = 0, has double root = 2. Therefore, the
general solution is
y = C1 te2t + C2 e2t
Problem 2. Fi
Quiz 5 Solution
Math 307 J. February 20, 2015
Problem 1. Find the general solution to the equation
y + 9y = (2t3 5) sin(t) + (2t3 6) sin(3t) + 5e3t t2 + 3t.
Do not nd undetermined coecients, just leave them as A, B, . . .
Solution. First, let us solve the