Steel Design 6th Edition Segui Solutions Manual

1. 1
DIFFERENTIAL EQUATIONS
(Problem Sets)
A. SEPARABLE DIFFERENTIAL EQUATIONS
Problem 1:
𝑑𝑦
𝑑𝑥
= 𝑥𝑦2
Solution: Analytical
𝑑𝑦
𝑑𝑥
= 𝑥𝑦2
𝑑𝑦
𝑦2
= 𝑥𝑑𝑥
𝑦−2
𝑑𝑦 = 𝑥𝑑𝑥
∫ 𝑦−2
𝑑𝑦 = ∫ 𝑥𝑑𝑥
(−
𝑦−1
1
) =
𝑥2
2
+
𝑐
2
−
1
𝑦
=
𝑥2
+ 𝑐
2
𝑦 = −
1
(
𝑥2 + 𝑐
2
)
𝑦 = −
1
(
𝑥2
2
) + 𝑐
Solution: MATLAB
Problem 2:
𝑑𝑦
𝑑𝑥
+ 𝑥𝑒𝑦
= 0
Solution: Analytical
𝑑𝑦
𝑑𝑥
= −𝑥𝑒𝑦

2. 2
𝑑𝑦
𝑒𝑦
= −𝑥𝑑𝑥
∫
𝑑𝑦
𝑒𝑦
= − ∫ 𝑥𝑑𝑥
𝑒−𝑦
=
𝑥2
2
+ 𝑐
ln⁡
(𝑒−𝑦
) = ln⁡
(
𝑥2
2
+ 𝑐)
−𝑦 = ln (
𝑥2
2
+ 𝑐)
𝑦 = − ln(
𝑥2
2
+ 𝑐)
Solution: MATLAB
B. EQUATIONS WITH HOMOGENEOUS COEFFICIENTS
Problem 1: 2(2𝑥2
+ 𝑦2
)𝑑𝑥 − 𝑥𝑦𝑑𝑦 = 0
Solution: Analytical
𝑙𝑒𝑡,
𝑦 = 𝑣𝑥
𝑑𝑦 = 𝑣𝑑𝑥 + 𝑥𝑑𝑣
⁡
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
2(2𝑥2
+ 𝑣2
𝑥2
)𝑑𝑥 − 𝑣𝑥2
(𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
4𝑥2
𝑑𝑥 + 2𝑣2
𝑥2
𝑑𝑥 − 𝑣2
𝑥2
𝑑𝑥 − 𝑣𝑥3
𝑑𝑣 = 0
4𝑥2
𝑑𝑥 + 𝑣2
𝑥2
𝑑𝑥 − 𝑣𝑥3
𝑑𝑣 = 0
𝑥2
(4 + 𝑣2
)𝑑𝑥 − 𝑣𝑥3
𝑑𝑣 = 0
𝑑𝑥
𝑥
−
𝑣𝑑𝑣
4 + 𝑣2
= 0
∫
𝑑𝑥
𝑥
− ∫
𝑣𝑑𝑣
4 + 𝑣2
= 0
ln⁡
(𝑥) −
1
2
ln⁡
(4 + 𝑣2) = ln⁡
(𝑐)
2ln⁡
(𝑥) − ln⁡
(4 + 𝑣2) = 2ln⁡
(𝑐)
ln⁡
(𝑥2
) − ln⁡
(4 + 𝑣2) = ln⁡
(𝑐2
)

3. 3
ln⁡
(𝑥2
) = ln⁡
(𝑐2
) + 𝑙𝑛(4 + 𝑣2)
ln⁡
(𝑥2
) = ln⁡
(𝑐2
)(4 + 𝑣2)
𝑥2
= 𝑐2
(4 + 𝑣2)
𝑓𝑟𝑜𝑚
𝑦 = 𝑣𝑥
𝑣 =
𝑦
𝑥
⁡
𝑡ℎ𝑢𝑠,
𝑥2
= 𝑐2
(4 +
𝑦2
𝑥2
)
𝑥2
= 𝑐2
(
4𝑥2
+ 𝑦2
𝑥2
)
𝑥4
= 𝑐2
(4𝑥2
+ 𝑦2
)⁡
Solution: MATLAB
Problem 2: 3(3𝑥2
+ 𝑦2
)𝑑𝑥 − 2𝑥𝑦𝑑𝑦 = 0
Solution: Analytical
𝑙𝑒𝑡,
𝑦 = 𝑣𝑥
𝑑𝑦 = 𝑣𝑑𝑥 + 𝑥𝑑𝑣
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
3(3𝑥2
+ 𝑣2
𝑥2)𝑑𝑥 − 2𝑣𝑥2(𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
3(3 + 𝑣2
)𝑥2
𝑑𝑥 − 2𝑣𝑥2(𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
3(3 + 𝑣2)𝑑𝑥 − 2𝑣(𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
9𝑑𝑥 + 3𝑣2
𝑑𝑥 − 2𝑣2
𝑑𝑥 − 2𝑣𝑥𝑑𝑣 = 0
9𝑑𝑥 + 𝑣2
𝑑𝑥 − 2𝑣𝑥𝑑𝑣 = 0
1
𝑥(9+𝑣2)
[(9 + 𝑣2)𝑑𝑥 − 2𝑣𝑥𝑑𝑣 = 0]

4. 4
𝑑𝑥
𝑥
−
2𝑣𝑑𝑣
9 + 𝑣2
= 0
𝑑𝑥
𝑥
=
2𝑣𝑑𝑣
9 + 𝑣2
∫
𝑑𝑥
𝑥
= ∫
2𝑣𝑑𝑣
9 + 𝑣2
ln(𝑥) = ln(9 + 𝑣2) + ln(𝑐)
ln(𝑥) − ln(9 + 𝑣2) = ln(𝑐)
ln(
𝑥
9 + 𝑣2
) = ln(𝑐)
𝑥
9 + 𝑣2
= 𝑐
𝑥 = 𝑐(9 + 𝑣2
)
𝑓𝑟𝑜𝑚,
𝑦 = 𝑣𝑥
𝑣 =
𝑦
𝑥
𝑡ℎ𝑢𝑠,
𝑥 = 𝑐 (9 +
𝑦2
𝑥2
)
𝑥 = 𝑐 (
9𝑥2
+ 𝑦2
𝑥2
)
𝑥3
= 𝑐(9𝑥2
+ 𝑦2
)
Solution: MATLAB
C. EXACT DIFFERENTIAL EQUATIONS
Problem 1: (𝑥 + 𝑦)𝑑𝑥 + (𝑥 − 𝑦)𝑑𝑦 = 0
Solution: Analytical

5. 5
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
𝜕𝑀
𝜕𝑦
=
𝑑
𝑑𝑦
(𝑥 + 𝑦)
𝜕𝑁
𝜕𝑥
=
𝑑
𝑑𝑥
(𝑥 − 𝑦)
⁡⁡⁡⁡⁡⁡⁡= 0 + 1 = 1 − 0
⁡⁡⁡⁡⁡⁡⁡= 1 = 1
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
= 1 ∴ 𝑜. 𝑑. 𝑒⁡𝑖𝑠⁡𝑒𝑥𝑎𝑐𝑡
𝜕𝐹
𝜕𝑥
= 𝑀(𝑥, 𝑦)
𝜕𝐹
𝜕𝑦
= 𝑁(𝑥, 𝑦)
∫ 𝑑𝑓 = ∫(𝑥 + 𝑦)𝑑𝑥 ∫ 𝑑𝑓 = ∫(𝑥 − 𝑦)𝑑𝑦
⁡𝑓 =
𝑥2
2
+ 𝑥𝑦 + 𝑄𝑦 𝑓 = 𝑥𝑦 −
𝑦2
2
+ 𝑅𝑥
𝑥2
2
+ 𝑥𝑦 + 𝑄𝑦 = ⁡𝑥𝑦 −
𝑦2
2
+ 𝑅𝑥
𝑅𝑥 =⁡
𝑥2
2
, 𝑄𝑦 = −
𝑦2
2
∴ ⁡⁡𝑓 =
𝑥2
2
+ 𝑥𝑦 −
𝑦2
2
⁡⁡⁡⁡⁡⁡𝑐 =
𝑥2
2
+ 𝑥𝑦 −
𝑦2
2
Solution: MATLAB
Problem 2: (6𝑥 + 𝑦2)𝑑𝑥 + 𝑦(2𝑥 − 3)𝑑𝑦 = 0
Solution: Analytical
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
𝜕𝑀
𝜕𝑦
=
𝑑
𝑑𝑦
(6𝑥 + 𝑦2)
𝜕𝑁
𝜕𝑥
=
𝑑
𝑑𝑥
𝑦(2𝑥 − 3)
⁡⁡⁡⁡⁡⁡⁡= 0 + 2𝑦 = 2𝑦 − 0

6. 6
⁡⁡⁡⁡⁡⁡⁡= 2𝑦 = 2𝑦
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
= 2𝑦 ∴ 𝑜. 𝑑. 𝑒⁡𝑖𝑠⁡𝑒𝑥𝑎𝑐𝑡
𝜕𝐹
𝜕𝑥
= 𝑀(𝑥, 𝑦)
𝜕𝐹
𝜕𝑦
= 𝑁(𝑥, 𝑦)
∫ 𝑑𝑓 = ∫(6𝑥 + 𝑦2)𝑑𝑥 ∫ 𝑑𝑓 = ∫ 𝑦(2𝑥 − 3)𝑑𝑦
⁡𝑓 = 3𝑥2
+ 𝑥𝑦2
+ 𝑄𝑦 𝑓 = 𝑥𝑦2
−
3𝑦2
2
+ 𝑅𝑥
3𝑥2
+ 𝑥𝑦2
+ 𝑄𝑦 = ⁡𝑥𝑦2
−
3𝑦2
2
+ 𝑅𝑥
𝑅𝑥 =⁡⁡3𝑥2
, 𝑄𝑦 = −
3𝑦2
2
∴ ⁡⁡𝑓 = 3𝑥2
+ 𝑥𝑦2
−
3𝑦2
2
⁡⁡⁡⁡⁡⁡𝑐 = 3𝑥2
+ 𝑥𝑦2
−
3𝑦2
2
Solution: MATLAB
D.LINEAR DIFFERENTIAL EQUATIONS
Problem 1: (5𝑥 + 3𝑦)𝑑𝑥 − 𝑥𝑑𝑦 = 0
Solution: Analytical
1
𝑥𝑑𝑥
[(5𝑥 + 3𝑦)𝑑𝑥 − 𝑥𝑑𝑦 = 0]
𝑑𝑦
𝑑𝑥
− 𝑥4
−
3𝑦
𝑥
= 0
𝑑𝑦
𝑑𝑥
−
3𝑦
𝑥
= 𝑥4

7. 7
𝑦𝑒−3 ∫
𝑑𝑥
𝑥 = ∫ 𝑥4
𝑒−3 ∫
𝑑𝑥
𝑥 𝑑𝑥
𝑦𝑒−3𝑙𝑛𝑥
= ∫ 𝑥4
𝑒−3𝑙𝑛𝑥
𝑑𝑥
𝑦𝑥−3
= ∫ 𝑥𝑑𝑥
𝑦𝑥−3
=
𝑥2
2
+ 𝑐
𝑦 =
𝑥5
2
+ 𝑐𝑥3
Solution: MATLAB
Problem 2:
𝑑𝑦
𝑑𝑥
= 𝑥 − 2𝑦
Solution: Analytical
𝑑𝑦
𝑑𝑥
= 𝑥 − 2𝑦
𝑑𝑦
𝑑𝑥
+ 2𝑦 = 𝑥
𝑦𝑒2 ∫ 𝑑𝑥
= ∫ 𝑥𝑒2 ∫ 𝑑𝑥
𝑑𝑥
𝑦𝑒2𝑥
= ∫ 𝑥𝑒2𝑥
𝑑𝑥
𝑦𝑒2𝑥
=
𝑥𝑒2𝑥
2
−
𝑒2𝑥
4
+ 𝑐
𝑦 =
𝑥
2
−
1
4
+ 𝑐𝑒−2𝑥
Solution: MATLAB

8. 8
E. ELEMENTARY APPLICATIONS OF DIFFERENTIAL
EQUATIONS
Problem 1: Consider a tank used in certain hydrodynamic experiments. After one
experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter.
To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at
a rate of 2liters/min, the well-stirred solution flowing out at the same rate. Find the time
that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution: Analytical
𝑑𝐿
𝑑𝑡
= −𝑘𝐿
𝑑𝐿
𝑑𝑡
= 𝑘𝐿⁡⁡⁡⁡
∫
𝑑𝐿
𝐿
= −𝑘 ∫ 𝑑𝑡 2 = 𝑘(200)
⁡⁡⁡⁡⁡⁡ ln⁡
(𝐿) = −𝑘𝑡 + 𝑐 𝑘 =
2
200
𝐿 = 𝑒−𝑘𝑡+𝑐
𝒌 = 𝟎. 𝟎𝟏
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ 𝐿 = 𝐶𝑒−𝑘𝑡
⁡𝑡 = 0, 𝑡 =? , 𝐿 = 2, 𝑘 = 0.01
𝐿 = 𝐶𝑒−𝑘𝑡
𝐿 = 𝐶𝑒−𝑘𝑡
⁡⁡⁡⁡⁡⁡⁡⁡ 200 = 𝐶𝑒0
2 = 200𝑒−0.01𝑡
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ 𝑪 = 𝟐𝟎𝟎 ln(0.01) = −0.01𝑡
𝑡 = −
ln⁡
(0.01)
0.01
𝑡 = 460.52⁡𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Solution: MATLAB
Problem 2: A thermometer is moved from room where the temperature is 70 F to a freezer
where the temperature is 12 F .After 30 seconds the thermometer reads 40 F. What does
it read after 2 minutes?

9. 9
Solution: Analytical
𝑇 = 𝑇𝑒 + 𝐶𝑒−𝑘𝑡
𝑡 = 0,
70 = 12 + 𝐶𝑒0
𝑪 = 𝟓𝟖
𝑡 = 0.5,
40 = 12 + 58𝑒−𝑘(0.5)
40 − 12 = ⁡58𝑒−0.5𝑘
28
58
= 𝑒−0.5𝑘
𝑘 =
− ln (
28
58
)
0.5
𝒌 = 𝟏. 𝟒𝟓𝟔𝟒𝟕𝟕
𝑡 = 2,
𝑇 = 12 + 58𝑒−(1.456477)(2)
𝑇 = 15.15⁡℉
Solution: MATLAB
F. INTEGRATING FACTORS FOUND BY INSPECTION
Problem 1: 𝑦(2𝑥𝑦 + 1)𝑑𝑥 − 𝑥𝑑𝑦 = 0
Solution: Analytical
𝑦(2𝑥𝑦 + 1)𝑑𝑥 − 𝑥𝑑𝑦 = 0
1
𝑦2
[2𝑥𝑦2
𝑑𝑥 + 𝑦𝑑𝑥 − 𝑥𝑑𝑦 = 0]
2𝑥𝑑𝑥 + 𝑑 (
𝑥
𝑦
) = 0
∫ 2𝑥𝑑𝑥 + ∫ 𝑑 (
𝑥
𝑦
) = 0
𝑥2
+
𝑥
𝑦
= 𝑐
𝑥2
𝑦 + 𝑥 = 𝑐𝑦

10. 10
Solution: MATLAB
Problem 2: 𝑦(𝑦2
+ 1)𝑑𝑥 + 𝑥(𝑦2
− 1)𝑑𝑦 = 0
Solution: Analytical
𝑦(𝑦2
+ 1)𝑑𝑥 + 𝑥(𝑦2
− 1)𝑑𝑦 = 0
𝑦3
𝑑𝑥 + 𝑦𝑑𝑥 + 𝑥𝑦2
𝑑𝑦 − 𝑥𝑑𝑦 = 0
1
𝑦2
[(𝑦𝑑𝑥 − 𝑥𝑑𝑦) + 𝑦2(𝑦𝑑𝑥 + 𝑥𝑑𝑦) = 0]
𝑑 (
𝑥
𝑦
) + 𝑑(𝑥𝑦) = 0
∫ 𝑑 (
𝑥
𝑦
) + ∫ 𝑑(𝑥𝑦) = 0
𝑥
𝑦
+ 𝑥𝑦 = 𝑐
𝑥 + 𝑥𝑦2
= 𝑐𝑦
𝑥(1 + 𝑦2) = 𝑐𝑦
Solution: MATLAB
G.DETERMINATION OF INTEGRATING FACTORS
Problem 1: 𝑦(4𝑥 + 𝑦)𝑑𝑥 − 2(𝑥2
− 𝑦)𝑑𝑦 = 0
Solution: Analytical
𝑦(4𝑥 + 𝑦)𝑑𝑥 − 2(𝑥2
− 𝑦)𝑑𝑦 = 0

11. 11
𝜕𝑀
𝜕𝑦
−
𝜕𝑁
𝜕𝑥
= 4𝑥 + 2𝑦 − (−4𝑥)
⁡⁡⁡= 4𝑥 + 2𝑦 + 4𝑥
= 8𝑥 + 2𝑦
1
𝑀
(
𝜕𝑀
𝜕𝑦
−
𝜕𝑁
𝜕𝑥
) =
2(4𝑥+𝑦)
𝑦(4𝑥+𝑦)
=
2
𝑦
𝐼. 𝐹
= 𝑒
−2 ∫
𝑑𝑦
𝑦
= 𝑒−2ln⁡
(𝑦)
= 𝑦−2
𝑦−2
[𝑦(4𝑥 + 𝑦)𝑑𝑥 − 2(𝑥2
− 𝑦)𝑑𝑦 = 0]
4𝑥𝑦−1
𝑑𝑥 + 𝑑𝑥 − 2𝑥2
𝑦−2
𝑑𝑦 + 2𝑦−1
𝑑𝑦 = 0
(4𝑥𝑦−1
𝑑𝑥 − 2𝑥2
𝑦−2
𝑑𝑦) + 𝑑𝑥 + 2𝑦−1
𝑑𝑦 = 0
(
4𝑥𝑑𝑥
𝑦
−
2𝑥2
𝑑𝑦
𝑦2
) + 𝑑𝑥 +
2𝑑𝑦
𝑦
= 0
2𝑑 (
𝑥2
𝑦
) + 𝑑𝑥 +
2𝑑𝑦
𝑦
= 0
2∫ 𝑑(
𝑥2
𝑦
) + ∫ 𝑑𝑥 + ∫
2𝑑𝑦
𝑦
= 0
2 (
𝑥2
𝑦
) + 𝑥 + 2 ln(𝑦) = 𝑐
2𝑥2
+ 𝑥𝑦 + 2𝑦 ln(𝑦) = 𝑐𝑦
Solution: MATLAB
Problem 2: (𝑥2
+ 𝑦2
+ 1)𝑑𝑥 + 𝑥(𝑥 − 2𝑦)𝑑𝑦 = 0
Solution: Analytical
(𝑥2
+ 𝑦2
+ 1)𝑑𝑥 + 𝑥(𝑥 − 2𝑦)𝑑𝑦 = 0
𝜕𝑀
𝜕𝑦
−
𝜕𝑁
𝜕𝑥
= 2𝑦 − (2𝑥 − 2𝑦)
⁡⁡⁡= 2𝑦 − 2𝑥 + 2𝑦

12. 12
= −2𝑥 + 4𝑦
= −2(𝑥 − 2𝑦)
1
𝑁
(
𝜕𝑀
𝜕𝑦
−
𝜕𝑁
𝜕𝑥
) =
−2(𝑥−2𝑦)
𝑥(𝑥−2𝑦)
= −
2
𝑥
𝐼. 𝐹
= 𝑒−2 ∫
𝑑𝑥
𝑥
= 𝑒−2ln⁡
(𝑥)
= 𝑥−2
𝑥−2[(𝑥2
+ 𝑦2
+ 1)𝑑𝑥 + 𝑥(𝑥 − 2𝑦)𝑑𝑦 = 0]
𝑑𝑥 + 𝑥−2
𝑦2
𝑑𝑥 + 𝑥−2
𝑑𝑥 + 𝑑𝑦 − 2𝑥−1
𝑦𝑑𝑦 = 0
𝑑𝑥 + 𝑥−2
𝑑𝑥 + 𝑑𝑦 +
𝑦2
𝑑𝑥
𝑥2
−
2𝑦𝑑𝑦
𝑥
= 0
𝑑𝑥 + 𝑥−2
𝑑𝑥 + 𝑑𝑦 +
𝑦2
𝑑𝑥 − 2𝑥𝑦𝑑𝑦
𝑥2
= 0
𝑑𝑥 + 𝑥−2
𝑑𝑥 + 𝑑𝑦 −
𝑥(2𝑦𝑑𝑦) − 𝑦2
𝑑𝑥
𝑥2
= 0
𝑑𝑥 + 𝑥−2
𝑑𝑥 + 𝑑𝑦 − 𝑑 (
𝑦2
𝑥
) = 0
∫ 𝑑𝑥 + ∫ 𝑥−2
𝑑𝑥 + ∫ 𝑑𝑦 − ∫ 𝑑 (
𝑦2
𝑥
) = 0
𝑥 −
1
𝑥
+ 𝑦 −
𝑦2
𝑥
= 0
𝑥2
− 1 + 𝑥𝑦 − 𝑦2
= 𝑐𝑥
𝑥2
− 𝑦2
+ 𝑥𝑦 − 1 = 𝑐𝑥
Solution: MATLAB
H.SUBSTITUTION SUGGESTED BY THE EQUATION
Problem 1:
𝑑𝑦
𝑑𝑥
= (9𝑥 + 4𝑦 + 1)2
Solution: Analytical

13. 13
𝑑𝑦
𝑑𝑥
= (9𝑥 + 4𝑦 + 1)2
𝑑𝑦 = (9𝑥 + 4𝑦 + 1)2
𝑑𝑥
𝑙𝑒𝑡,
𝑢 = 9𝑥 + 4𝑦 + 1
𝑑𝑢 = 9𝑑𝑥 + 4𝑑𝑦
𝑑𝑦 =
1
4
(𝑑𝑢 − 9𝑑𝑥)
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
1
4
(𝑑𝑢 − 9𝑑𝑥) = (𝑢)2
𝑑𝑥
𝑑𝑢 − 9𝑑𝑥 = 4𝑢2
𝑑𝑥
𝑑𝑢 = 4𝑢2
𝑑𝑥 + 9𝑑𝑥
𝑑𝑢 = (4𝑢2
+ 9)𝑑𝑥
𝑑𝑢
(4𝑢2 + 9)
= 𝑑𝑥
∫
𝑑𝑢
(4𝑢2 + 9)
= ∫ 𝑑𝑥
1
2
[
1
3
arctan (
2𝑢
3
)] = 𝑥 +
1
6
𝑐
1
6
arctan (
2𝑢
3
) = 𝑥 +
1
6
𝑐
arctan (
2𝑢
3
) = 6𝑥 + 𝑐
2𝑢
3
= tan⁡
(6𝑥 + 𝑐)
2𝑢 = 3 tan(6𝑥 + 𝑐)
𝑓𝑟𝑜𝑚,
𝑢 = 9𝑥 + 4𝑦 + 1
2(9𝑥 + 4𝑦 + 1) = 3 tan(6𝑥 + 𝑐)
18𝑥 + 8𝑦 = 3 tan(6𝑥 + 𝑐)
Solution: MATLAB

14. 14
Problem 2:
𝑑𝑦
𝑑𝑥
= sin⁡
(𝑥 + 𝑦)
Solution: Analytical
𝑑𝑦
𝑑𝑥
= sin(𝑥 + 𝑦)
𝑑𝑦 = sin(𝑥 + 𝑦)𝑑𝑥
𝑙𝑒𝑡,
𝑢 = 𝑥 + 𝑦
𝑑𝑢 = 𝑑𝑥 + 𝑑𝑦
𝑑𝑦 = 𝑑𝑢 − 𝑑𝑥
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛,
𝑑𝑢 − 𝑑𝑥 = sin(𝑢) 𝑑𝑥
𝑑𝑢 = sin(𝑢) 𝑑𝑥 + 𝑑𝑥
𝑑𝑢 = (sin(𝑢) + 1)𝑑𝑥
𝑑𝑢
sin(𝑢) + 1
= 𝑑𝑥
𝑑𝑢
sin(𝑢) + 1
(
1 − sin(𝑢)
1 − sin(𝑢)
) = 𝑑𝑥
(1 − sin(𝑢))𝑑𝑢
1 − sin2(𝑢)
= 𝑑𝑥
(1 − sin(𝑢))𝑑𝑢
cos2(𝑢)
= 𝑑𝑥
𝑑𝑢
cos2(𝑢)
−
sin(𝑢) 𝑑𝑢
cos2(𝑢)
= 𝑑𝑥
∫ sec2(𝑢)𝑑𝑢 − ∫ cos−2(𝑢) (sin(𝑢))𝑑𝑢 = ∫ 𝑑𝑥
tan(𝑢) − cos−1(𝑢) = 𝑥 + 𝑐

15. 15
tan(𝑢) −
1
cos(𝑢)
= 𝑥 + 𝑐
tan(𝑢) − sec(𝑢) = 𝑥 + 𝑐
𝑓𝑟𝑜𝑚,
𝑢 = 𝑥 + 𝑦
tan(𝑥 + 𝑦) − sec(𝑥 + 𝑦) = 𝑥 + 𝑐
Solution: MATLAB
I. BERNOULLI'S EQUATION
Problem 1:
𝑑𝑦
𝑑𝑥
= 𝑦 − 𝑥𝑦3
𝑒−2𝑥
Solution: Analytical
𝑑𝑦
𝑑𝑥
= 𝑦 − 𝑥𝑦3
𝑒−2𝑥
𝑑𝑦
𝑑𝑥
− 𝑦 = −𝑥𝑒−2𝑥
𝑦3
𝑑𝑦 − 𝑦𝑑𝑥 = −𝑥𝑒−2𝑥
𝑦3
𝑑𝑥
𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖′
𝑠⁡𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛,
𝑑𝑦 + 𝑃𝑦𝑑𝑥 = 𝑄𝑦𝑛
𝑑𝑥
𝑓𝑟𝑜𝑚⁡𝑤ℎ𝑖𝑐ℎ,
𝑃 = −1
𝑄 = −𝑥𝑒−2𝑥
𝑛 = 3
(1 − 𝑛) = −2
𝑣 = 𝑦1−𝑛
= 𝑦−2
𝐼. 𝐹,
𝑢 = 𝑒(1−𝑛) ∫ 𝑃𝑑𝑥

16. 16
𝑢 = 𝑒−2 ∫(−1)𝑑𝑥
𝑢 = 𝑒2 ∫ 𝑑𝑥
𝑢 = 𝑒2𝑥
𝑣𝑢 = (1 − 𝑛) ∫ 𝑄𝑢𝑑𝑥 + 𝑐
𝑦−2
(𝑒2𝑥
) = (−2) ∫(−𝑥𝑒−2𝑥)(𝑒2𝑥)𝑑𝑥 + 𝑐
𝑦−2
𝑒2𝑥
= 2 ∫ 𝑥𝑑𝑥 + 𝑐
𝑒2𝑥
𝑦2
= 2 (
𝑥2
2
) + 𝑐
𝑒2𝑥
= 𝑦2
(𝑥2
+ 𝑐)
Solution: MATLAB
Problem 2: 𝑥2 𝑑𝑦
𝑑𝑥
− 𝑦2
= 2𝑥𝑦
Solution: Analytical
𝑥2
𝑑𝑦
𝑑𝑥
− 𝑦2
= 2𝑥𝑦
𝑑𝑦 −
2𝑦
𝑥
𝑑𝑥 =
𝑦2
𝑥2
𝑑𝑥
𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖′
𝑠⁡𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛,
𝑑𝑦 + 𝑃𝑦𝑑𝑥 = 𝑄𝑦𝑛
𝑑𝑥
𝑓𝑟𝑜𝑚⁡𝑤ℎ𝑖𝑐ℎ,
𝑃 = −
2
𝑥
𝑄 =
1
𝑥2

17. 17
𝑛 = 2
(1 − 𝑛) = −1
𝑣 = 𝑦1−𝑛
= 𝑦−1
𝐼. 𝐹,
𝑢 = 𝑒(1−𝑛) ∫ 𝑃𝑑𝑥
𝑢 = 𝑒2 ∫
𝑑𝑥
𝑥
𝑢 = 𝑒2ln⁡
(𝑥)
𝑢 = 𝑥2
𝑣𝑢 = (1 − 𝑛) ∫ 𝑄𝑢𝑑𝑥 + 𝑐
𝑦−1
(𝑥2
) = (−1) ∫ (
1
𝑥2
)(𝑥2)𝑑𝑥 + 𝑐
𝑥2
𝑦−1
= − ∫ 𝑑𝑥 + 𝑐
𝑥2
𝑦
= −𝑥 + 𝑐
𝑦 = 𝑥2
(−𝑥 + 𝑐)−1
Solution: MATLAB
J. COEFFICIENTS LINEAR IN THE TWO VARIABLES
Problem 1: (𝑥 + 𝑦 − 1)𝑑𝑥 + (2𝑥 + 2𝑦 + 1)𝑑𝑦 = 0
Solution: Analytical
(𝑥 + 𝑦 − 1)𝑑𝑥 + (2𝑥 + 2𝑦 + 1)𝑑𝑦 = 0
1
(2𝑥 + 2𝑦 + 1)𝑑𝑥
[(𝑥 + 𝑦 − 1)𝑑𝑥 + (2𝑥 + 2𝑦 + 1)𝑑𝑦 = 0]

18. 18
𝑥 + 𝑦 − 1
2𝑥 + 2𝑦 + 1
+
𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
= −
𝑥 + 𝑦 − 1
2𝑥 + 2𝑦 + 1
𝑙𝑒𝑡,
𝑢 = 𝑥 + 𝑦
𝑑𝑢 = 𝑑𝑥 + 𝑑𝑦
𝑑𝑢
𝑑𝑥
= 1 +
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
− 1
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
(
𝑑𝑢
𝑑𝑥
− 1) = −
(𝑢) − 1
2(𝑢) + 1
𝑑𝑢
𝑑𝑥
= −
𝑢 − 1
2𝑢 + 1
− 1
𝑑𝑢
𝑑𝑥
= −
𝑢 − 1
2𝑢 + 1
−
2𝑢 + 1
2𝑢 + 1
𝑑𝑢
𝑑𝑥
=
𝑢 + 2
2𝑢 + 1
(2𝑢 + 1)𝑑𝑢
𝑢 + 2
= 𝑑𝑥
∫
(2𝑢 + 1)𝑑𝑢
𝑢 + 2
= ∫ 𝑑𝑥
𝑙𝑒𝑡,
𝑣 = 𝑢 + 2
𝑑𝑣 = 𝑑𝑢
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
∫
(2(𝑣 − 2) + 1)𝑑𝑣
𝑣
= ∫ 𝑑𝑥
∫
(2𝑣 − 3)𝑑𝑣
𝑣
= ∫ 𝑑𝑥
2𝑣 − 3 ln(𝑣) = 𝑥 + 𝑐
𝑓𝑟𝑜𝑚,
𝑣 = 𝑢 + 2
2(𝑢 + 2) − 3 ln(𝑢 + 2) = 𝑥 + 𝑐
𝑓𝑟𝑜𝑚,
𝑢 = 𝑥 + 𝑦

19. 19
2(𝑥 + 𝑦 + 2) − 3 ln(𝑥 + 𝑦 + 2) = 𝑥 + 𝑐
2𝑥 + 2𝑦 + 4 − 3 ln(𝑥 + 𝑦 + 2) = 𝑥 + 𝑐
2𝑥 − 𝑥 + 2𝑦 − 3 ln(𝑥 + 𝑦 + 2) = 𝑐
𝑥 + 2𝑦 − 3 ln(𝑥 + 𝑦 + 2) = 𝑐
Solution: MATLAB
Problem 2: (𝑥 − 𝑦 + 2)𝑑𝑥 + 3𝑑𝑦 = 0
Solution: Analytical
(𝑥 − 𝑦 + 2)𝑑𝑥 + 3𝑑𝑦 = 0
1
3𝑑𝑥
[(𝑥 − 𝑦 + 2)𝑑𝑥 + 3𝑑𝑦 = 0]
𝑥 − 𝑦 + 2
3
+
𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
= −
𝑥 − 𝑦 + 2
3
𝑙𝑒𝑡,
𝑢 = 𝑥 − 𝑦
𝑑𝑢 = 𝑑𝑥 − 𝑑𝑦
𝑑𝑢
𝑑𝑥
= 1 −
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 1 −
𝑑𝑢
𝑑𝑥
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
(1 −
𝑑𝑢
𝑑𝑥
) = −
𝑢 + 2
3
𝑑𝑢
𝑑𝑥
= 1 +
𝑢 + 2
3
𝑑𝑢
𝑑𝑥
=
𝑢 + 5
3
∫
3𝑑𝑢
𝑢 + 5
= ∫ 𝑑𝑥
3 ln(𝑢 + 5) = 𝑥 + 𝑐

20. 20
𝑓𝑟𝑜𝑚,
𝑢 = 𝑥 − 𝑦
3 ln(𝑥 − 𝑦 + 5) = 𝑥 + 𝑐
Solution: MATLAB
K. HOMOGENEOUS LINEAR EQUATIONS
Problem 1: 𝑦′′
+ 𝑦 = 0, 𝑦(0) = 2, 𝑦′(0) = 3
Solution: Analytical
𝑦′′
+ 𝑦 = 0
𝑚2
+ 1 = 0
𝑚 = ±𝑖
𝑦(𝑥) = 𝐶1 cos(𝑥) + 𝐶2sin⁡
(𝑥)
𝑦′
(𝑥) = −𝐶1sin⁡
(𝑥) + 𝐶2 cos(𝑥)
𝑦(0) = 2
2 = 𝐶1 cos(0) + 𝐶2 sin(0)
2 = 𝐶1 + 0
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐶1 = 2
𝑦′(0) = 3
3 = −𝐶1sin⁡
(0) + 𝐶2 cos(0)
3 = 0 + 𝐶2
𝐶2 = 3
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒⁡⁡𝐶1𝑎𝑛𝑑⁡𝐶2,
𝑦(𝑥) = 2 cos(𝑥) + 3 sin(𝑥)
Solution: MATLAB

21. 21
Problem 2: 𝑦′′
− 6𝑦′
+ 8𝑦 = 0, 𝑦(0) = 1, 𝑦′(0) = 6
Solution: Analytical
𝑦′′
− 6𝑦′
+ 8𝑦 = 0
𝑚2
− 6𝑚 + 8 = 0
𝑚 = 2, 4
𝑦(𝑥) = 𝐶1𝑒2𝑥
+ 𝐶2𝑒4𝑥
𝑦′
(𝑥) = 2𝐶1𝑒2𝑥
+ 4𝐶2𝑒4𝑥
𝑦(0) = 1
1 = 𝐶1𝑒0
+ 𝐶2𝑒0
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐶1 + 𝐶2 = 1 𝑒𝑞. 1
𝑦′(0) = 6
6 = 2𝐶1𝑒0
+ 4𝐶2𝑒0
2𝐶1 + 4𝐶2 = 6 𝑒𝑞. 2
𝑒𝑞𝑢𝑎𝑡𝑒⁡𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛⁡1⁡𝑎𝑛𝑑⁡2, 𝑤𝑒⁡𝑔𝑒𝑡
𝐶1 = −1
𝐶2 = 2
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒⁡⁡𝐶1𝑎𝑛𝑑⁡𝐶2,
⁡𝑦(𝑥) = −𝑒2𝑥
+ 2𝑒4𝑥

22. 22
Solution: MATLAB
L. NONHOMOGENEOUS LINEAR EQUATIONS
Problem 1: 𝑦′′
− 3𝑦′
− 4𝑦 = 30𝑒𝑥
Solution: Analytical
𝑦′′
− 3𝑦′
− 4𝑦 = 30𝑒𝑥
𝑚2
− 3𝑚 − 4 = 0
(𝑚 − 4)(𝑚 + 1) = 0
𝑚 = 4, −1
𝑚′
= 1
𝑦𝑐 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
𝑦𝑝 = 𝐴𝑒𝑥
𝑦𝑝
′
= 𝐴𝑒𝑥
𝑦𝑝
′′
= 𝐴𝑒𝑥
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
(𝐴𝑒𝑥) − 3(𝐴𝑒𝑥) − 4(𝐴𝑒𝑥) = 30𝑒𝑥
𝑓𝑎𝑐𝑡𝑜𝑟⁡𝑜𝑢𝑡⁡𝑒𝑥
,
𝐴 − 3𝐴 − 4𝐴 = 30
−6𝐴 = 30
𝐴 = −5
𝑡ℎ𝑢𝑠,
𝑦 = 𝑦𝑐 + 𝑦𝑝
𝑦 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
+ 𝐴𝑒𝑥
⁡𝑦 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
− 5𝑒𝑥

23. 23
Solution: MATLAB
Problem 2: 𝑦′′
− 3𝑦′
− 4𝑦 = 30𝑒4𝑥
Solution: Analytical
𝑦′′
− 3𝑦′
− 4𝑦 = 30𝑒𝑥
𝑚2
− 3𝑚 − 4 = 0
(𝑚 − 4)(𝑚 + 1) = 0
𝑚 = 4, −1
𝑚′
= 4
𝑦𝑐 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
𝑦𝑝 = 𝐴𝑥𝑒4𝑥
𝑦𝑝
′
= 𝐴𝑒4𝑥
+ 4𝐴𝑥𝑒4𝑥
𝑦𝑝
′′
= 8𝐴𝑒4𝑥
+ 16𝐴𝑥𝑒4𝑥
𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒,
(8𝐴𝑒4𝑥
+ 16𝐴𝑥𝑒4𝑥) − 3(𝐴𝑒4𝑥
+ 4𝐴𝑥𝑒4𝑥) − 4(𝐴𝑥𝑒4𝑥) = 30𝑒𝑥
𝑓𝑎𝑐𝑡𝑜𝑟⁡𝑜𝑢𝑡⁡𝑒4𝑥
,
8𝐴 + 16𝐴𝑥 − 3𝐴 − 4𝐴𝑥 = 30
8𝐴 + 0 − 3𝐴 − 0 = 30
5𝐴 = 30
𝐴 = 6
𝑡ℎ𝑢𝑠,
𝑦 = 𝑦𝑐 + 𝑦𝑝
𝑦 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
+ 𝐴𝑥𝑒4𝑥
⁡𝑦 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
+ 6𝑥𝑒4𝑥
Solution: MATLAB

24. 24