HW 0 Math 225
Review practice exercises
For each question, submit only problems a), c), e)
Due ------
1. Evaluate the integral by using the substitution method

a) ∫ cos 3
b) ∫ (4 + 2)10
c) ∫ 2 √3 + 1
d) ∫ 2(3 + 2)4
e) ∫(2 ? )6
f) ∫ 3 √24 ? 1
g) ∫ 5 √3 + 1
3

h) ∫

√1+2

4
0

2. Evaluate the integral by using the integration by parts method.

a) ∫ cos 5
b) ∫ 2?
c) ∫ 2 cos
d) ∫ 2 sin 3
e) ∫
ln
2

2
1

f) ∫

2

1
0


3. Trigonometric integrals and Trigonometric substitutions.

a)∫ (sin )5 (cos )3
3 4?
2?


b) ∫ (cos )5
2?
0


c) ∫(1 + cos )2

d) ∫
3
√16?2

2√3
0


e) ∫ 3√2 + 4
2
0



4. Evaluate the integral using partial fractions method.

a) ∫
?9
(+5)(?2)

b) ∫
2+2?1
3?

c) ∫
2
(?3)(+2)2

d) ∫
4+1
(2+1)2

e) ∫
+1
(2+1)2

5. Operations with vectors and their geometry
a) Consider the vectors = [
2
1
] = [
1
3
]. Evaluate + , 2 + 3, 2 ? and illustrate your results graphically
using the parallelogram law for addition of vectors.
b) Consider the vectors = [
1
2
3
] = [
?2
4
5
]. Evaluate ? = = (the dot product or inner product of .
c) Given the vector = [
1
2
3
], find its magnitude, and find all vectors that are perpendicular to .
d) Draw the set of solutions of the equation 1 + 22 = 3 and also the set of solutions of the equation ?1 + 2 = 4
e) Find numbers such that [
1
?1
] + [
2
1
] is equal to [
3
4
]. Illustrate this geometrically.

We can use MATLAB to solve this problem numerically on a given interval.

Let’s choose the interval [0,3]
tspan = [0 3]; % Find solution on this interval
yprime = @(t,y) -y + 4*exp(t); % Define the DE
y0 = 3; % Set the initial condition
[t,y] = ode45(yprime,tspan,y0); % Use solver ode45 to get the solution
plot(t,y) % Display results


Lecture 5. (Wednesday September 01) Review
Practice exercises


Homework 1
Due (Thursday September 02)
Quiz 2 covers this material

Section 1.1
4, 6, 9, 10
Section 1.2
3, 6, 11, 15, 19, 23, 33, 40, 44
Section 1.3
5, 13, 15, 31
Section 1.4
2, 4, 12, 16, 22
Section 1.5
4, 6, 9
Section 1.6
1, 5, 10, 24


For the following problems, submit only the even numbers.

Determine the order of the given differential equation; also state whether the equation is linear or nonlinear.

1.

+ 2 = 0

2.
2
2
+ cos ( + ) =

Verify that each given function is a solution of the differential equation.

3. ′ ? = 2 ; = 3 + 2

4. 22′′ + 3′ ? = 0, > 0 ; 1() =
1
2 , 2() = ?1

Determine the value of for which the given differential equation has a solution of the form = .

5. ′ + 2 = 0

6. ′′′ ? 3′′ + 2′ = 0

Determine the value of for which the given differential equation has a solution of the form = for > .

7. 2′′ + 4′ + 2 = 0

8. 2′′ ? 4′ + 4 = 0

Verify that () satisfies the given differential equation. Then determine a value of the constant so that ()
satisfies the given initial condition.
9. ′ + (sin ) = 0 ; () = cos , () = 1
10. ′ + 2 = 0; () = ?2, (0) = 1
Basic Methods to solve DE.
Solve the following problems using the Separation of variables Method.

11. 2
2
1 y
x
dx
dy
?=

12. ,

13.

14.

15. ,
16. Solve the initial value problem , and determine the interval in which the solution
is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.

17. Solve the initial value problem , and determine where the solution attains its
maximum value.

Solve the following problems using the Integrating Factor Method.

18.
19. ,
20. ,
)1(2
243 2
?
++=
y
xx
dx
dy 1)0( ?=y
3
3
4
4
y
xx
dx
dy
+
?=
y
x
ey
ex
dx
dy
+
?=
?
y
ee
dx
dy xx
43+
?=
?
1)0( =y
yy
x
dx
dy
63
31
2
2
?
+= 1)0( =y
)23(
2cos2
y
x
dx
dy
+= 1)0( ?=y
tyy ?=?c 42
242 tyyt =+c 2)1( =y
22 =+c tyy 1)0( =y
21.
22.
23.
24. , ,
25. Consider the initial value problem
,
a) Describe the behavior of the solution for large .
b) Determine the value of for which the solution first intersects the line .
26. Show that if and are positive constants, and is any real number, then every solution of the equation
has the property that as . Hint: Consider the case and separately.

)sin(0 tkAkTkuu Z+=+c
2
22 ttetyy ?=+c
)2sin(5 tyy =+c
)sin(2 tyyt =+c 1)( 2 =Sy 0!t
)2cos(2341 tyy +=+c 0)0( =y
t
t 12=y
a O b
tbeayy O?=+c 0→y f→t O=a Oza