MIDTERM EXAM REVIEW PROBLEMS
DUE NEVER.
1. Equations of first order
Solve the following initial value problems. If no initial value is given, find the general
solution. If you feel like it, draw a direction field and plot some of the solutions.
(1) ty 0 + 2y = sin(t),
y(π/2) = 1,
(2) ty 0 + (t + 1)y = t,
t>0
y(ln(2)) = 1,
(3) y 0 (3 + 4y) = e−x − ex ,
t>0
y(0) = 1
(4) (2xy 2 + 2y) + (2x2 y + 2x)y 0 = 0
(5) t3 y 0 + 4t2 y = e−t ,
y(−1) = 0,
t<0
(6) (3x2 − 2xy + 2)dx + (6y 2 − x2 + 3)dy = 0
(7) y 0 = (ay + b)/(cy + d),
a, b, c, d constants.
(8) sin(t)y 0 + cost(t)y = et ,
y(1) = a,
(9) y 0 − y = 1 + 3 sin(t),
(10) y 0 = x(x2 + 1)/4y 3 ,
0<t<π
y(0) = 0,
√
y(0) = −1/ 2
(11) (ex sin(y) − 2y sin(x))dx + (ex cos(y) + 2 cos(x))dy = 0
2
(12) y 0 + 2ty = 2te−t
(13) y 0 = 2x/(y + x2 y),
y(0) = −2
(14) (3x2 y + 2xy + y 3 )dx + (x2 + y 2 )dy = 0
(15) y 0 = e2x + y − 1
1
2
DUE NEVER.
2. Equations of second order
Solve the following initial value problems. If no initial value is given, find the general
solution. When solutions are given, they are solutions to the associated homogeneous
equation.
(1) y 00 − 2y 0 − 8y = 0
(2) t2 y 00 − t(t + 2)y 0 + (t + 2)y = 0,
(3) y 00 − 6y 0 + 82y = 0,
t > 0,
y(0) = −1,
y1 (t) = t
y 0 (0) = 2
(4) 9y 00 + 9y 0 − 4y = 0
(5) xy 00 − y 0 + 4x3 y = 0
x > 0,
y1 (x) = sin(x)
(6) y 00 + 2y 0 + 5y = 4e−t cos(2t),
(7) y 00 + 4y 0 + 5y = 0,
y(0) = 1,
y(0) = 1,
y 0 (0) = 0
y 0 (0) = 0
(8) y 00 − 2y 0 + y = et /(1 + t2 )
(9) 2y 00 + y 0 − 4y = 0,
(10) y 00 + y = tan(t),
y(0) = 0,
y 0 (0) = 1
0 < t < π/2
(11) y 00 + y = 3 sin(2t) + t cos(2t)
(12) y 00 − 2y 0 − 3y = 3te2t ,
y(0) = 1,
(13) ty 00 − (1 + t)y 0 + y = t2 e2t ,
y 0 (0) = 0
t > 0, y1 (t) = (1 + t),
y2 (t) = et
3. Series solutions
Use power series techniques to solve the following second order equations by providing
a recurrence relation to determine all the coefficients of the power series near zero. When
possible, find the general term. Classify any singular points as regular or irregular.
(1) (2 + x2 )y 00 − xy 0 + 4y = 0
(2) x2 y 00 − 5xy 0 + 9y = 0
(3) (1 + x2 )y 00 − 4xy 0 + 6y = 0
MIDTERM EXAM REVIEW PROBLEMS
(4) 2x2 y 00 + 3xy 0 + (2x2 − 1)y = 0
(5) (1 − x)y 00 + xy 0 − y = 0
(6) x2 y 00 − x(3 + x)y 0 + (x + 3)y = 0
(7) x(x − 1)y 00 + 6x2 y 0 + 3y = 0
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