1. The document provides analytical solutions and MATLAB code for various types of differential equations, including separable, homogeneous, exact, and linear differential equations. 2. Applications include models for thermometer cooling, concentration of dye in a mixing tank, and hydrodynamic experiments in a tank. 3. Methods demonstrated include finding integrating factors by inspection or determining integrating factors when a differential equation is not exact.

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: