Steel Design 6th Edition Segui Solutions Manual

1. 1
Problem Set with MATLAB Solution
SEPARABLE DIFFERENTIAL EQUATIONS
1..
𝑑𝑦
𝑑𝑥
= 𝑥𝑦2
Analytical:
𝑑𝑦
𝑦2
= 𝑥𝑑𝑥
𝑦−2𝑑𝑦 = 𝑥𝑑𝑥
∫ 𝑦−2𝑑𝑦 = ∫ 𝑥𝑑𝑥
𝑦−1
1
=
𝑥2
2
+
𝑐
2
𝑦 = −
1
𝑥2 + 𝑐
2
𝒚 = −
𝟏
𝒙𝟐
𝟐
+𝒄
Matlab:

2. 2
2.
𝑑𝑦
𝑑𝑥
+ 𝑥𝑒𝑦
= 0
Analytical:
𝑑𝑦
𝑑𝑥
− 𝑥𝑒𝑦
∫
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)
Matlab:

3. 3
EQUATIONS WITH HOMOGENEOUS COEFFICIENTS
1. 3(3𝑥2 + 𝑦2)𝑑𝑥 − 2𝑥𝑦𝑑𝑦 = 0
Analytical:
𝐿𝑒𝑡 𝑦 = 𝑣𝑥, 𝑣 =
𝑦
𝑥
𝑑𝑦 = 𝑣𝑑𝑥 + 𝑥𝑑𝑣
3(3𝑥2
+ 𝑣2
𝑥2)𝑑𝑥− 2𝑣𝑥2(𝑣𝑑𝑥+ 𝑥𝑑𝑣) = 0
3(3 + 𝑣2) 𝑥2 𝑑𝑥 − 2𝑣𝑥2 (𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
9𝑑𝑥 + 3𝑣2𝑑𝑥 − 2𝑣2𝑑𝑥 − 2𝑣𝑥𝑑𝑣 = 0
9𝑑𝑥 + 𝑣2𝑑𝑥 − 2𝑣𝑥𝑑𝑣 = 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
)
𝑥 = 𝑐(
9 + 𝑣2
𝑥2
)
𝑥 = 𝑐(9 +
𝑣2
𝑥2
)
𝑥3
= 𝑐(9𝑥2
+ 𝑦2
)

4. 4
Matlab:
2. 2(2𝑥2
+ 𝑦2
)𝑑𝑥 − 𝑥𝑦𝑑𝑦 = 0
Analytical:
𝐿𝑒𝑡 𝑦 = 𝑣𝑥, 𝑣 =
𝑦
𝑥
𝑑𝑦 = 𝑣𝑑𝑥 + 𝑥𝑑𝑣
2(2𝑥2 + 𝑣2𝑥2) 𝑑𝑥 − 𝑣𝑥2 (𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
4𝑥2𝑑𝑥 + 2𝑣2𝑥2𝑑𝑥 − 𝑣2𝑥2𝑑𝑥 − 𝑣𝑥3𝑑𝑣 = 0
𝑥2 (4 + 𝑣2) 𝑑𝑥 − 𝑣𝑥3𝑑𝑣 = 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)
ln(𝑥2
) = ln(𝑐2
) + ln(4 + 𝑣2
)
ln(𝑥2
) = ln(𝑐2
)(4 + 𝑣2
)

5. 5
𝑒ln(𝑥2
)
= 𝑒 ln(𝑐2
)(4+ 𝑣2
)
𝑥2
= (𝑐2
)(4 + 𝑣2
)
𝑥2
= 𝑐2
(4 +
𝑦2
𝑥2
)
𝑥2
= 𝑐2
(
4 + 𝑦2
𝑥2
)
𝒙𝟒
= 𝒄𝟐
(𝟒𝒙𝟐
+ 𝒚𝟐
)
Matlab:

6. 6
EXACT DIFFERENTIAL EQUATIONS
1. (𝑥 + 𝑦)𝑑𝑥 + (𝑥 − 𝑦)𝑑𝑦 = 0
Analytical:
Test for exactness
𝜕𝑀
𝜕𝑦
= 𝑥 + 𝑦
𝜕𝑀
𝜕𝑦
= 𝑥 − 𝑦
𝜕𝑀
𝜕𝑦
= 1
𝜕𝑀
𝜕𝑦
= 1
∫ 𝑑𝐹 = ∫(𝑥 + 𝑦)𝑑𝑥
𝐹 =
𝑥2
2
+ 𝑥𝑦 + 𝑄(𝑦)
∫ 𝑑𝐹 = ∫(𝑥 − 𝑦)𝑑𝑦
𝐹 = 𝑥𝑦 −
𝑦2
2
+ 𝑅(𝑥)
𝑥2
2
+ 𝑥𝑦 + 𝑄(𝑦) = 𝑥𝑦 −
𝑦2
2
+ 𝑅(𝑥)
𝑄(𝑦) = −
𝑦2
2
, 𝑅(𝑥) =
𝑥2
2
𝑪 =
𝒙𝟐
𝟐
+ 𝒙𝒚 = −
𝒚𝟐
𝟐
Matlab:

7. 7
2. (6𝑥 + 𝑦2)𝑥 + 𝑦(2𝑥 − 3)𝑑𝑦 = 0
Analytical:
Test for exactness
𝜕𝑀
𝜕𝑦
= 6𝑥 + 𝑦2
𝜕𝑀
𝜕𝑦
= 𝑦(2𝑥 − 3)
𝜕𝑀
𝜕𝑦
= 2𝑦
𝜕𝑀
𝜕𝑦
= 2𝑦
∫ 𝑑𝐹 = ∫(6𝑥 + 𝑦2)𝑑𝑥
𝐹 = 3𝑥2
+ 𝑥𝑦2
+ 𝑄(𝑦)
∫ 𝑑𝐹 = ∫ 𝑦(2𝑥 − 3) 𝑑𝑦
𝐹 = 𝑥𝑦2
−
3𝑦2
2
+ 𝑅(𝑥)
3𝑥2
+ 𝑥𝑦2
+ 𝑄(𝑦) = 𝑥𝑦2
−
3𝑦2
2
+ 𝑅(𝑥)
𝑄(𝑦) = −
3𝑦2
2
, 𝑅(𝑥) = 3𝑥2
𝟑𝒙𝟐
+ 𝒙𝒚𝟐
−
𝟑𝒚𝟐
𝟐
Matlab:

8. 8
LINEAR DIFFERENTIAL EQUATIONS
1. (5𝑥 + 3𝑦)𝑑𝑥 − 𝑥𝑑𝑦 = 0
Analytical:
𝑑𝑦
𝑑𝑥
− 𝑥−4
−
3𝑦
𝑥
= 0
𝑑𝑦
𝑑𝑥
−
3𝑦
𝑥
= 𝑥4
𝑦𝑒
−3 ∫
𝑑𝑥
𝑥 = ∫ 𝑥4
𝑒
−3 ∫
𝑑𝑥
𝑥 𝑑𝑥
𝑦𝑒−3 ln(𝑥)
= ∫ 𝑥4
𝑒−3 ln(𝑥)
𝑑𝑥
𝑦𝑥−3 = ∫ 𝑥𝑑𝑥
𝑦𝑥−3 =
𝑥2
2
+ 𝑐
𝑦 =
𝑥5
2
+ 𝑐𝑥3
Matlab:

9. 9
2.
𝑑𝑦
𝑑𝑥
= 𝑥 − 2𝑦
Analytical:
𝑑𝑦
𝑑𝑥
+ 2𝑦 = 𝑥
𝑦𝑒2∫ 𝑑𝑥
= ∫𝑥 𝑒2 ∫ 𝑑𝑥
𝑑𝑥
𝑦𝑒2𝑥
= ∫ 𝑥 𝑒2𝑥
𝑦𝑒2𝑥
=
𝑥𝑒2𝑥
2
−
𝑒2𝑥
4
+ 𝑐
𝒚 =
𝒙
𝟐
−
𝟏
𝟒
+ 𝒄𝒆−𝟐𝒙
Matlab:

10. 10
APPLICATIONS OF DIFFERENTIAL EQUATIONS
1. 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?
Analytical:
𝑇 = 𝑇𝑒 + 𝐶𝑒
−𝑘𝑡
𝑤ℎ𝑒𝑛 𝑡 = 0
70 = 12𝐶𝑒
0
𝑪 = 𝟓𝟖
𝑤ℎ𝑒𝑛 𝑡 = 0.5
40 = 12 + 58𝑒−0.5𝑘
40 − 12 = 58𝑒−0.5𝑘
28
58
= 𝑒−0.5𝑘
𝒌 = 𝟏.𝟒𝟓𝟔𝟒𝟕𝟕
𝑤ℎ𝑒𝑛 𝑡 = 2
𝑇 = 12 + 58𝑒−1.456477(𝑐)
𝑻 = 𝟏𝟓. 𝟏𝟓℃
Matlab:

11. 11
2. 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.
Analytical:
𝑑𝐿
𝑑𝑡
= −𝑘𝐿
𝑑𝐿
𝑑𝑡
= −𝑘𝐿
𝑑𝐿
𝐿
= −𝑘 ∫𝑑𝑡
𝑙𝑛(𝐿) = −𝑘𝑡 + 𝑐
𝐿 = 𝑒−𝑘𝑡+𝑐
𝐿 = 𝐶𝑒
−𝑘𝑡
𝑤ℎ𝑒𝑛 𝑡 = 0
𝐶𝑒
0
= 200
𝑪 = 𝟐𝟎𝟎
Matlab:

12. 12
INTEGRATING FACTORS FOUND BY INSPECTION
1. 𝑦(2𝑥𝑦 + 1)𝑑𝑥 − 𝑥𝑑𝑦 = 0
Analytical:
(2𝑥𝑦 + 1)𝑑𝑥 − 𝑥𝑑𝑦 = 0
2𝑥𝑑𝑥 + 𝑑 (
𝑥
𝑦
) = 0
∫2𝑥𝑑𝑥 + ∫ 𝑑 (
𝑥
𝑦
) = 0
𝑥2 +
𝑥
𝑦
= 𝑐
𝒄𝒚 = 𝒙𝟐
𝒚 + 𝒙
Matlab:
2. (𝑦2 + 1)𝑑𝑥 + 𝑥(𝑦2 − 1)𝑑𝑦 = 0
Analytical:
𝑦3𝑑𝑥 + 𝑦𝑑𝑥 + 𝑥𝑦2𝑑𝑦 − 𝑥𝑑𝑦 = 0
∫𝑑 (
𝑥
𝑦
) + ∫𝑥𝑦 = 0
𝑥
𝑦
+ 𝑥𝑦 = 𝑐𝑦
(𝟏 + 𝒚𝟐) = 𝒄𝒚
Matlab:

13. 13
DETERMINATION OF INTEGRATING FACTORS
1. (𝑦 − 𝑥𝑦)𝑑𝑥 + 𝑥𝑑𝑦 = 0
Analytical:
Test for exactness:
𝜕𝑀
𝜕𝑦
= 1 − 𝑥 ;
𝜕𝑁
𝜕𝑥
= 1
1
𝑥
[(1 − 𝑥) − 1] = 𝑓(𝑥)
−1 = 𝑓(𝑥)
𝐼𝐹 = 𝑒∫𝑓(𝑥)𝑑𝑥 = 𝑒−𝑥
(𝑦𝑒−𝑥 − 𝑥𝑦𝑒−𝑥)𝑑𝑥+ 𝑥𝑒−𝑥𝑑𝑦 = 0
∫𝜕𝐹 = ∫(𝑦𝑒−𝑥 − 𝑥𝑦𝑒−𝑥)𝑑𝑥
𝐹 = 𝑦 ∫𝑒−𝑥 − 𝑦 ∫𝑥𝑒−𝑥
𝐿𝑒𝑡 𝑢 = 𝑥 ; 𝑑𝑣 = 𝑒−𝑥𝑑𝑥
𝑑𝑢 = 𝑑𝑥 𝑣 = −𝑒−𝑥
𝐹 = 𝑥𝑦𝑒−𝑥 + 𝑄(𝑦)
∫𝜕𝐹 = ∫(𝑥𝑒−𝑥)𝑑𝑦
𝐹 = 𝑥𝑦𝑒−𝑥 + 𝑅(𝑥)
𝑥𝑦𝑒−𝑥 + 𝑄(𝑦) = 𝑥𝑦𝑒−𝑥 + 𝑅(𝑥)
𝑄(𝑦) = 0 , 𝑅(𝑥) = 0
𝑪 = 𝒙𝒚𝒆−𝒙
Matlab:
2. 𝑦(𝑦 + 2𝑥 − 2)𝑑𝑥 − 2(𝑥 + 𝑦)𝑑𝑦 = 0

14. 14
Analytical:
Test for exactness:
𝜕𝑀
𝜕𝑦
= 2𝑦 + 2𝑥 − 2 ;
𝜕𝑁
𝜕𝑥
= −2
2𝑦 + 2𝑥 − 2 + 2
−(2𝑥 + 2𝑦)
= 𝑓(𝑥)
−1 = 𝑓(𝑥)
𝐼𝐹 = 𝑒∫ 𝑓(𝑥)𝑑𝑥 = 𝑒−𝑥
(𝑒−𝑥
𝑦2
+ 2𝑒−𝑥
𝑥𝑦 − 2𝑒−𝑥
𝑦)𝑑𝑥
−(2𝑒−𝑥
𝑥 + 2𝑒−𝑥
𝑦)𝑑𝑦 = 0
∫ 𝜕𝐹 = ∫ (𝑒
−𝑥
𝑦2 + 2𝑒−𝑥𝑥𝑦− 2𝑒−𝑥𝑦)𝑑𝑥
𝐹 = −𝑒−𝑥𝑦2 − 2𝑒−𝑥𝑥𝑦 + 𝑄(𝑦)
∫ 𝜕𝐹 = ∫−(2𝑒−𝑥𝑥 + 2𝑒−𝑥𝑦)𝑑𝑦
𝐹 = −2𝑒−𝑥𝑥𝑦 − 𝑒−𝑥𝑦2 + 𝑅(𝑥)
−𝑒−𝑥
𝑦2
− 2𝑒−𝑥
𝑥𝑦 + 𝑄(𝑦)
= −2𝑒−𝑥𝑥𝑦 − 𝑒−𝑥𝑦2 + 𝑅(𝑥)
𝑄(𝑦) = 0 , 𝑅(𝑥) = 0
𝑪𝒆𝒙 = 𝒚(𝒚 + 𝟐𝒙)
Matlab:

15. 15
SUBSTITUTION SUGGESTED BY THE EQUATION
1.
𝑑𝑦
𝑑𝑥
= (9𝑥 + 4𝑦 + 1)2
Analytical:
𝑑𝑦 = (9𝑥 + 4𝑦 + 1)2
𝑑𝑥
𝐿𝑒𝑡 𝑦 = 9𝑥 + 4𝑦 + 1
𝑑𝑦 = 9𝑑𝑥 + 4𝑑𝑦
1
4
(𝑑𝑢 − 9𝑑𝑥) = 𝑢2
𝑑𝑥
𝑑𝑢 − 9𝑑𝑥 = 4𝑢2
𝑑𝑥
∫
𝑑𝑢
4𝑢2 + 9
= ∫ 𝑑𝑥
arctan(
2𝑢
3
) = 6𝑥 + 𝑐
2𝑢 = 3tan(6𝑥 + 𝑐)
2(9𝑥 + 4𝑦 + 1) = 3tan(6𝑥 + 𝑐)
𝟏𝟖𝒙+ 𝟖𝒚 = 𝟑𝐭𝐚𝐧(𝟔𝒙 + 𝒄)
Matlab:

16. 16
2.
𝒅𝒚
𝒅𝒙
= 𝒔𝒊𝒏(𝒙 + 𝒚)
Analytical:
𝑑𝑦 = 𝑠𝑖𝑛(𝑥 + 𝑦)𝑑𝑥
𝐿𝑒𝑡 𝑢 = 𝑥 + 𝑦 , 𝑑𝑢 = 𝑑𝑥 + 𝑑𝑦
𝑑𝑢 − 𝑑𝑥 = sin(𝑢) 𝑑𝑥
𝑑𝑢 = sin(𝑢)𝑑𝑥 + 𝑑𝑥
𝑑𝑢
sin(𝑢) + 1
= 𝑑𝑥
𝑑𝑢
sin(𝑢) + 1
(
1 − sin(𝑢)
1 − sin(𝑢)
) = 𝑑𝑥
(1 − sin(𝑢))
𝑐𝑜𝑠2(𝑢)
𝑑𝑢 = 𝑑𝑥
∫ 𝑠𝑒𝑐2(𝑢)𝑑𝑢 − ∫ 𝑐𝑜𝑠−2(𝑢) sin(𝑢)𝑑𝑢 = ∫ 𝑑𝑥
tan(𝑢) − sec(𝑢) = 𝑥 + 𝑐
𝐭𝐚𝐧(𝒙 + 𝒚) − 𝐬𝐞𝐜(𝒙 + 𝒚) = 𝒙 + 𝒄
Matlab:

17. 17
BERNOULLI'S EQUATION
1. 𝑦(6𝑦2
− 𝑥 − 1)𝑑𝑥 + 2𝑥𝑑𝑦 = 0
Analytical:
𝑑𝑦
𝑑𝑥
+
3𝑦3
𝑥
−
𝑦
2
−
𝑦
2𝑥
= 0
𝑦−3𝑑𝑦
𝑑𝑥
+ (−
1
2
−
1
2𝑥
)(𝑦−2) = −
3
𝑥
𝐿𝑒𝑡 𝑣 = 𝑦−2 , 𝑑𝑣 = −2𝑦−3𝑑𝑦
𝑑𝑣
𝑑𝑥
+ (−
1
2
−
1
2𝑥
) (−2𝑣) =
6
𝑥
𝑑𝑣
𝑑𝑥
+ (1 +
1
𝑥
) 𝑣 =
6
𝑥
𝑃 = 1 +
1
𝑥
; 𝑄 =
6
𝑥
𝑣𝑒
∫(1+
1
𝑥
)𝑑𝑥 = 6
𝑥
𝑒
∫(1+
1
𝑥
)𝑑𝑥
𝑣𝑥𝑒𝑥 = 6𝑒𝑥 + 𝑐
𝑦−2𝑥𝑒𝑥 = 6𝑒𝑥 + 𝑐
𝒙𝒆𝒙
= 𝟔𝒚𝟐
𝒆𝒙
+ 𝒄𝒚𝟐
Matlab:

18. 18
2. . 𝑦′ = 𝑦 − 𝑥𝑦3𝑒−2𝑥
Analytical:
𝑑𝑦 − 𝑦 𝑑𝑥 = −𝑥𝑒−2𝑥𝑦3𝑑𝑥
𝐹𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ:
𝑃 = −1 (1 − 𝑛) = −1
𝑄 = −𝑥𝑒−𝑥 𝑧 = 𝑦1−𝑛
𝑥 = 3 = 𝑦−2
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟:
𝑧𝑢 = (1 − 𝑛)∫𝑄𝑢𝑑𝑥+ 𝑐
𝑦−2(𝑒2𝑥) = −2 ∫(−𝑥𝑒−2𝑥)(𝑒2𝑥)𝑑𝑥 + 𝑐
𝑒2𝑥𝑦−2 = 2 ∫𝑥𝑑𝑥 + 𝑐
𝑒2𝑥
𝑦2
= 𝑥2 + 𝑐
𝒆𝟐𝒙 = 𝒚𝟐(𝒙𝟐 + 𝒄)
Matlab:

19. 19
COEFFICIENTS LINEAR IN THE TWO VARIABLES
1. (𝑥 − 𝑦 + 2)𝑥 + 3𝑑𝑦 = 0
Analytical:
(𝑥 − 𝑦 + 2)𝑑𝑥 + 3𝑑𝑦 = 0
𝑑𝑦
𝑑𝑥
+
𝑥 − 𝑦 + 2)
3
= 0
𝑑𝑦
𝑑𝑥
= −
𝑥 − 𝑦 + 2
3
𝐿𝑒𝑡 𝑢 = 𝑥 − 𝑦
𝑑𝑢 = 𝑑𝑥 − 𝑑𝑦
𝑑𝑢
𝑑𝑥
= 1 −
𝑑𝑦
𝑑𝑥
(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
𝟑 𝒍𝒏(𝒙 − 𝒚 + 𝟓) = 𝒙 + 𝒄
Matlab:

20. 20
2. (𝑥 + 𝑦 − 1)𝑑𝑥 + (2𝑥 + 2𝑦 + 1)𝑑𝑦 = 0
Analytical:
𝐿𝑒𝑡 𝑢 = 𝑥 + 𝑦
𝑑𝑢 = 𝑑𝑥 + 𝑑𝑦
𝑑𝑢
𝑑𝑥
= 1 +
𝑑𝑦
𝑑𝑥
(
𝑑𝑢
𝑑𝑥
− 1) = −
𝑢 − 1
2𝑢 + 1
𝑑𝑢
𝑑𝑥
= −
𝑢 − 1
2(𝑢) + 1
− 1
𝑑𝑢
𝑑𝑥
=
𝑢 − 1
2𝑢 + 1
+
2𝑢 + 1
2𝑢 + 1
𝑑𝑢
𝑑𝑥
=
𝑢 + 1
2𝑢 + 1
∫
(𝑢 + 1)
2𝑢 + 1
= ∫ 𝑑𝑥 𝐿𝑒𝑡 𝑣 = 𝑢 + 2 ,𝑑𝑣 = 𝑑𝑢
2𝑣 − 3 𝑙𝑛(𝑣) = 𝑥 + 𝑐
2(𝑢 + 2) − 3 𝑙𝑛(𝑢 + 2) = 𝑥 + 𝑐
2(𝑥 + 𝑦 + 2) − 3 𝑙𝑛(𝑥 + 𝑦 + 2) = 𝑥 + 𝑐
2𝑥 − 𝑥 + 2𝑦 − 3 𝑙𝑛(𝑥 + 𝑦 + 2) = 𝑐
𝒙 + 𝟐𝒚 − 𝟑 𝒍𝒏(𝒙 + 𝒚 + 𝟐) = 𝒄
Matlab:

21. 21
HOMOGENEOUS LINEAR EQUATIONS
1. y′′ − 6y′ + 8y = 0,y(0) = 1,y′(0) = 6
Analytical:
𝑚2
− 6𝑚 + 8 = 0
𝑚 = 2, 4
𝑦(𝑥) = 𝐶1𝑒2𝑥
+ 𝐶2𝑒4𝑥
𝑦 ′ (𝑥) = 2𝐶1e2𝑥
+ 4𝐶2e4𝑥
𝑤ℎ𝑒𝑛 𝑦(0) = 1
1 = 𝐶1𝑒0 + 𝐶2𝑒0
𝑪𝟏 + 𝑪𝟐 = 𝟏 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏
𝑤ℎ𝑒𝑛 𝑦 ′(0) = 6
6 = 2𝐶1𝑒0 + 𝐶2𝑒0
𝟐𝑪𝟏 + 𝟒𝑪𝟐 = 𝟔 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐
𝐸𝑄𝑈𝐴𝑇𝐸 1 𝐴𝑁𝐷 2
𝑆𝑜 𝑤𝑒 𝑔𝑒𝑡, 𝑪𝟏 = −1 ,𝑪𝟐 = 𝟐
𝒚 (𝒙) = −𝐞𝟐𝒙
+ 𝟐𝐞𝟒𝒙
Matlab:

22. 22
2. 𝑦 ′′ + 𝑦 = 0,𝑦(0) = 2,𝑦 ′(0) = 3
Analytical:
𝑚2
+ 1 = 0
𝑦(𝑥) = 𝐶1cos(𝑥) + 𝐶2sin(𝑥)
𝑦 ′ (𝑥) = 𝐶1𝑠𝑖𝑛(𝑥) + 𝐶2cos(𝑥)
𝑤ℎ𝑒𝑛 𝑦(0) = 1
2 = 𝐶1 cos(0) + 𝐶2sin(0)
𝟐 = 𝑪𝟏
𝑤ℎ𝑒𝑛 𝑦 ′(0) = 3
3 = 𝐶1 sin(0) + 𝐶2cos(0)
𝟑 = 𝑪𝟐
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝐶1 𝑎𝑛𝑑 𝐶2
𝒚(𝒙) = 𝟐 𝒄𝒐𝒔(𝒙) + 𝟑 𝒔𝒊𝒏(𝒙)
Matlab:

23. 23
NONHOMOGENEOUS LINEAR EQUATIONS
1. 𝑦 ′′ − 3𝑦 ′ − 4𝑦 = 30𝑒𝑥
Analytical:
𝑚2
− 3 − 4 = 0
(𝑚 − 4)(𝑚 + 1) = 0
𝑚 = 4,−1 , 𝑚′ = 1
𝑦𝑐 = 𝐶1𝑒4𝑥
+ 𝐶2𝑒−𝑥
𝑦𝑝 = 𝐴𝑒𝑥
𝑦′𝑝 = 𝐴𝑒𝑥
𝑦′′𝑝 = 𝐴𝑒𝑥
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒
(𝐴𝑒𝑥)− 3(𝐴𝑒𝑥) − 4(𝐴𝑒𝑥) = 30𝑒𝑥
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑒𝑥
𝐴 = −5
𝑦 = 𝑦𝑐 + 𝑦𝑝
𝑦 = 𝐶1𝑒4𝑥 + 𝐶2𝑒−𝑥 + 𝐴𝑒𝑥
𝒚 = 𝑪𝟏𝒆𝟒𝒙 + 𝑪𝟐𝒆−𝒙 − 𝟓𝒆𝒙
Matlab:

24. 24
2: 𝑦 ′′ − 3𝑦 ′ − 4𝑦 = 30𝑒𝑥
Analytical:
𝑚2 − 3 − 4 = 0
(𝑚 − 4)(𝑚 + 1) = 0
𝑚 = 4,−1 , 𝑚′ = 1
𝑦𝑐 = 𝐶1𝑒4𝑥 + 𝐶2𝑒−𝑥
𝑦𝑝 = 𝐴𝑥𝑒4𝑥
𝑦′𝑝 = 𝐴𝑥𝑒4𝑥 + 4𝐴𝑥𝑒4𝑥
𝑦′′𝑝 = 8𝑥𝑒4𝑥 + 16𝐴𝑥𝑒4𝑥
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒
(8𝑥𝑒4𝑥 + 16𝐴𝑥𝑒4𝑥)− 3(𝐴𝑒4𝑥 + 4𝐴𝑥𝑒4𝑥)− 4𝐴𝑥𝑒4𝑥 = 30𝑒𝑥
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑒4𝑥
𝐴 = 6
𝑦 = 𝑦𝑐 + 𝑦𝑝
𝑦 = 𝐶1𝑒4𝑥 + 𝐶2𝑒−𝑥 + 𝐴𝑒𝑥
𝒚 = 𝑪𝟏𝒆𝟒𝒙 + 𝑪𝟐𝒆−𝒙 + 𝟔𝒙𝒆𝟒𝒙
Matlab: