Engineering Calculus
Homework # 2
Homework # 2
7.1 # 4: (3 pts) (a) For what values of k does the function y = cos kt satisfy the differential
equation 4y ′′ = −25y?
(b) For those values of k, verify that every member of the family of functions y = A sin kt +
B cos kt is also a solution.
(a): (2 pts) Assume that y = cos kt is a solution to the equation and then
y ′ = −k sin kt
First derivative
y ′′ = −k 2 cos kt
y ′′ = −25y
Second derivative
Original equation
−4k 2 cos kt = −25 cos kt
Substituting for y
2
−4k = −25
Dividing by cos kt (assuming cos kt ̸= 0
k = ± 25
Solving for k
(b): (1 pt) Assume that a solution is a function like y = A sin kt + B cos kt with k = ± 52 ,
and test that it is a solution:
y ′ = ± 25 A cos(± 25 t) ∓ 25 B sin(± 25 t)
′′
2
25
2
y = − 25
4 A sin(± 5 t) − 4 B cos(± 5 t)
′′
y = −25y
)
(
)
( 25
2
25
2
4 − 4 A sin(± 5 t) − 4 B cos(± 5 t) = −25 A sin(± 52 t) + B cos(± 25 t)
(
(
)
−25 A sin(± 52 t) + B cos(± 25 t) ≡ −25 A sin(± 52 t) + B cos(± 25 t)
First derivative
Second derivative
Original equation
Substituting for y
7.2 # 21: (4 pts) Use Euler’s method with step size 0.5 to compute the approximate y-values
y1 , y2 , y3 , and y4 of the solution of the initial-value problem y ′ = y − 2x, y(1) = 0.
n
0
1
2
3
4
xn
1.0
1.5
2.0
2.5
3.0
yn
0.00
-1.00
-3.00
-6.50
-12.25
F (x, y)
-2.00
-4.00
-7.00
-11.50
1
2 F (x, y)
-1.00
-2.00
-3.50
-5.75
7.3 # 12: (3 pts) Find the solution of the differential equation that satisfies the given initial
dy
ln x
condition:
=
, y(1) = 2
dx
xy
1
Engineering Calculus
Homework # 2
dy
ln x
=
dx
xy
ln x
dx
y dy =
∫
∫x
ln x
y dy =
dx
x
∫
1 2
udu
2y =
Original equation
Separating variables
Integrating
Substituting u = ln x
= 21 u2 + C
= 12 (ln x)2 + C
√
y = ± (ln x)2 + C
√
2 = + (ln 1)2 + C
C=4
√
y = (ln x)2 + 4
The general solution
Using initial condition to find C (only “+” works)
The value for C
The full solution
2