Download Review Exercises:

First-Order Differential Equations
Review Exercises: - Chapter 1
Section 1.1 General and Particular Solution
Solve the following 1st order differential equation and obtain its general and particular solution.
(1)
dy
 3x 2  5
dx
(3)
dy
 x2  x ex x  0 , y  0
dx

x  1, y  1

(5) f ' ( x) 
x
f (0)  2 ,
4  x4
(2)
dy
 e2x
dx
x  0, y 
5
2
(4)
dy
 sin 5 x
dx
x  18 , y  20
(6)
d
f ( x)  ln x f (0)  1
dx
Section 1.2 To solve First-Order Differential Equations
Solve the following differential equation by separating the variable:
dy y 2  xy 2

(2)
dx x 2 y  x 2
dy
xy
(1)

dx 1  y
(3) ( y  3)
(5)
(4)
dy
2y

dx x( y  1)
(7) tan x
(9)
dy 4 y

dx
x
dy
 xy  y
dx
(6) cos 2 x

dy
 1 y
dx
dy
 3 x 2 ( y  2)
dx
dy
 y3
dx

(8) cos x e 2 y  y y'  e y sin 2 x
y (0)  8
(10) x
dy
 2y  2  x
dx
y (1)  0
Solve the following homogeneous equation using reducible to separable form
y
(11) y '  1 
x
put y  ux
y  y
(12) y '  1    
x x
y
(13) x
dy
 y ex
dx
put y  ux
(14) 2 x 2
dy
 x2  y2
dx
2
put y  ux
put y  ux
1
(15) ( x  y ) y ' x  y  0
(17) ( x  y  1)
put y  ux
dy
 (2 x  2 y  3)
dx
dy
 y 2  xy
dx
put y  ux
put u  x  y
(18) (2 x  4 y  5)dy  (2 y  x  2.5)dx
(19) dy  ( y  x  1) 2 dx
(16) x( x  y )
put u  x  2 y
put u  y  x  1
Section 1.3 Exact Differential Equations
Solve the following using exact differential equation:
(1) y 3 dx  3xy2 dy  0
(2) (2 x  e y )dx  xe y dy  0
(3) (2 xy 2  3)dx  (2 x 2 y  4)dy  0
(4) ( x 2  y 2 )dx  2 xy dy  0
(5) (5x  4 y)dx  4( x  2 y 3 )dy  0
(6) (3  y )dx  (cos y  x) dy  0
(7) 2xyex dx  e x dy  0
2
(9)
(11)

2
(8) e x y dx  (1  e x  y )dy  0
dy
yx y
 xy 2  cos x sin x
dx
dy
y cos xy  e 2 y

(10)
dx 2 xe2 y  x cos xy  2 y
dy
(3 x 2 y 2  e y )  2( xy 3  2)
dx
(12) (1  xe y )
2

dy
 ey
dx
Solve the following by determining the integrating factor:
(13)  y dx  x dy  0
(14) 3 dx  e y  x dy  0
(15) 2 y dx  xdy  0
(16) ( x  y 2 ) dx  xy dy  0
(17) 2 y dx  (1  6 x  y ) dy  0
(18)
(19) 2 y 2 dx  (1  8xy) dy  0
y 2 dx  3xy dy  0
(20)
y sin xdx  (2 cos x  4 y 2 )dy  0
Hint* (13-16) apply case (a), (17-20) apply case (b).
Section 1.4 Solving by Inspection Method
2
Solve the following 1st order linear differential equation by inspection method:





(3) 6 x 2 y  12 xy  y 2 dx  6 x 2  xy dy  0




(5) y 4  2 y dx  xy3  2 y 4  4 x dy  0
x  y  dx  x  y  dy  0
(7)


(2) xdy  y  x 2 e x dx
(1) xdy   ydx


(4) xdy  ydx  xy3 dy  0


(6) xdy  ydx 1  x 2  0
(8) ( x 2  3 y 2 ) dx  2 xydy  0

(9) 2 xy  2 y 2 dx  x 2  4 xy dy  0
(10) 3xdy  4 ydx  0
Section 1.5 First-Order Linear Differential Equations
Solve the following 1st order linear differential equation by using the integrating factor:
(1)
dy
 yx
dx
(3)
dy y
 x
dx x
(5) x
dy
 y  x sin x
dx
(7) ( x  1)
(9) ( x  1)
(2) y '5 y  e 2 x
, y (0)  0
(4)
(6) sin x
dy
 y  ( x  1) 2
dx
dy
 y  ( x  1) 4 ,
dx
dy  1 
   1 y  e x
dx  x 
dy
 y cos x  sin x cos x
dx
(8) (1  x 2 )
y (4)  9
(10) x
dy
 xy  2
dx
dy
 y  x cos x ,
dx
y ( )  0
Section 1.6 Bernoulli’s Equation
Solve the following Bernoulli’s equations:
(1)
dy 1
 y  xy 2
dx x
(3)
dy 1
 y  2x 2 y 3
dx x
(2)
dy
2x
 xy 
dx
y
(4) 3 y  2
dy
 y 4 e3x
dx
3
(5)
dy
 2 y  y 3  x  1
dx
(6) x 2 y 
1 3 dy
x
 y 3 cos x
2 dx
Section 1.7 Ricatti’s Equation
Solve the following Ricatti’s equations:
(1) x 2
dy
 2  2 xy  x 2 y 2  0
dx
dy
(3)
 y 2  2 xy  x 2  1
dx
(2) x
dy
 y  2 y 2  2x2
dx
dy y 2 y
(4)

 1
dx x 2 x
dy
dy
(6)
 y2  2 y  8  0
 y 2  xy  1
dx
dx
Section 1.8 Applications to series Electrical Circuit
(5) 2 x
(1) A simple R-L series circuit with R =1  , L =25 H and E (t )  e t V was connected.
Determine the current i(t) given that the initial current i(0) = 0.
(2) A 30 volts electromotive force is applied to an R-L series circuit which the resistance R is 2
ohms and inductance L is 25 Henry. Given the initial condition where i(0) = 0, find the current
i(t). Determine the current as t approaches infinity t   .
(3) A simple R-C circuit with R = 20  , C = 10 3 F and E (t )  e t V was connected. Determine
the charge q(t) given that the initial charge q(0) = 0. Determine the current i(t).
(4) A 100 volts electromotive force is applied to an R-C series circuit which the resistance R is
1000 ohms and capacitance C is 1  10 4 Farad. Given the initial condition where i(0) = 0.2,
find the charge q(t) on the capacitor. Determine the charge and current as t = 0.001 sec. If t
approaches infinity t   , find the charge q () .
4
Second-Order Ordinary Differential Equations
Review Exercises: - Chapter 2
Section 2.2 Constant Coefficient of Second Order Homogeneous Equation
Solve the general solution:
(problems 1-6: apply real roots, problems 7-10: apply equal roots, problems 11-16: apply
complex roots.)
(1)
d2y
dy
 7  12 y  0
2
dx
dx
d2y
dy
(3)
 4  5y  0
2
dx
dx
(2)
d2y
dy
 3  2y  0
2
dx
dx
d2y
dy
(4) 4 2  10  6 y  0
dx
dx
(5)
d2y
dy 3
2  y 0
2
dx 4
dx
(6)
d2y
dy
 9  9y  0
2
dx
dx
(7)
d2y
dy
 8  16 y  0
2
dx
dx
(8)
d2y
dy
 14  49 y  0
2
dx
dx
(9)
1 d2y
dy 3
 3
 y0
2
2 dx
dx 2
(10) 4
(11)
d2y
dy
 4  7y  0
2
dx
dx
d2y
dy
(13) 3 2  9  54 y  0
dx
dx
(15)
d2y
 64 y  0
dx 2
(12)
d2y
192 dy

 3y  0
2
4 dx
dx
d2y
dy
 2  10 y  0
2
dx
dx
d 2 y 4 dy 2
(14) 2 2 
 y0
3 dx 3
dx
(16)
d2y
 11y  0
dx 2
Solve the initial-value problems:
(17)
d 2 y dy

 2 y  0 , y ( 0 )  4 , y ' ( 0 )  5
dx 2 dx
(18)
d2y
dy
 4  5 y  0 , y (0)  0 , y ' (0)  2
2
dx
dx
5
(19)
d2y
dy
 4  4 y  0 , y (0)  1, y ' (0)  1
2
dx
dx
(20)
d2y
dy
 4  4 y  0 , y (0)  3 , y ' (0)  1
2
dx
dx
d2y
dy
 3.2  2.56 y  0 , y (0)  0 , y ' (0)  2
2
dx
dx
(21)
(22)
d2y
dy
 4  5 y  0 , y ( 0)  2 , y ' ( 0)  4
2
dx
dx
(23) 2
d2y
dy
 4  3 y  0 , y (0)  2 , y ' (0)  1
2
dx
dx
(24)
d2y
 13 y  0 , y (0)  1 , y ' (0)  3
dx 2
(25)
d2y
 3 y  0 , y ( 0 )  1 , y ' ( 0 )  3
dx 2
Solve the boundary-value problems:
(26)
d2y
dy
 4  5 y  0 , y (1)  0 , y ' (1)  1
2
dx
dx
(27)
d2y
dy
2 2
 2 y  0 , y (1)  2 , y ' (1)  0
2
dx
dx
(28)
d2y
dy
 
 2  5 y  0 , y ( 0)  1 , y    1
2
dx
dx
4
Section 2.3 Wronskian Test For Linear Independence of Solutions
(29) Using Wronskian equation, referring to problems (1, 2, 7, 8, 11, 12) and test for
independence:
Section 2.4 Non-homogeneous Equations
Solve the below problems (30 - 50) using Undetermined Coefficient Methods:
6
(30)
d2y
 4 y  8x 2
2
dx
(31)
d2y
 4 y  4x 2  2x  3
2
dx
(32)
d2y
dy
 5  6 y  12
2
dx
dx
(33)
d2y
dy
 5  6y  x2  2
2
dx
dx
(34)
d2y
dy
 5  6 y  4 sin 4 x
2
dx
dx
(35)
d2y
dy
 6  10 y  2 sin 2 x  2 cos 2 x
2
dx
dx
(36)
d2y
dy
 4  2 y  2x 2  x  2
2
dx
dx
d2y
dy
(37)
 3  2 y  2e x
2
dx
dx
(hint* apply rule 1: multiply by x)
(38)
d2y
dy
 2  3 y  4e x
2
dx
dx
(hint* apply rule 1: multiply by x)
(39)
d 2 y dy

 2 y  6e x
dx 2 dx
(hint* apply rule 1: multiply by x)
d2y
dy
(40)
 3  2 y  8e  x
2
dx
dx
(hint* apply rule 1: multiply by x)
(41)
d2y
dy
 2  y  3e  x
2
dx
dx
(hint* apply rule 1: multiply by x 2 )
(42)
d2y
dy
 8  16 y  5 e 4 x
2
dx
dx
(hint* apply rule 1: multiply by x 2 )
d2y
dy
(43)
 2  3 y  x  2e 2 x (hint* apply rule 2)
2
dx
dx
7
(44)
d2y
 4 y  4 x  5  6 xe2 x
2
dx
(45)
d2y
dy
 6  9 y  3x 2  1  6e 3 x (hint* apply rule 1 and 2 together)
2
dx
dx
(hint* apply rule 2)
Solve the initial value problems:
(46)
d2y
dy
1
3
 3  2 y  2 x 2 , y (0)  , y ' 0  
2
2
2
dx
dx
(47)
d2y
dy
 4  3 y  10e 2 x , y (0)  1 , y' 0  3
2
dx
dx
(48)
d2y
dy
7
 4  5 y  39e 3 x , y (0)  , y' 0  1
2
2
dx
dx
(49)
d2y
dy
1
5
 2  8 y  4e 2 x , y (0)  , y ' 0  
2
2
3
dx
dx
(50)
d 2 y dy

 2 y  e x  sin 2 x , y (0)  1 , y' 0  3
2
dx
dx
Solve the below problems (51-56) using variation of parameters:
(51)
d2y
dy
 3  2 y  e4x
2
dx
dx
(52)
d2y
dy
 6  9 y  xe3 x
2
dx
dx
(53)
d2y
 4 y  csc 2 x
dx 2
(54)
d2y
 4 y  tan 2 x
dx 2
d2y
dy
(55)
 3  2y  ex 1
2
dx
dx
8
(56)
d2y
dy
1
 3  2y  x
2
dx
dx
e 1
Section 2.5 Euler-Cauchy Differential Equations
Make use of either solution of the form y  x m or solution by Operator D method, solve the
following differential equation:
(57) x 2
d2y
dy
 x  8y  0
2
dx
dx
(58) x 2
d2y
dy
x y0
2
dx
dx
(59) x 2
d2y
dy
 x  9y  0
2
dx
dx
(60) x 2
d2y
dy
 3x  3 y  2 x 2
2
dx
dx
d2y
dy
(61) x
 5x  2 y  x 2
2
dx
dx
2
(62) x 2
d2y
dy
 5 x  4 y  2 sin 2 x
2
dx
dx
9