Republic of the Philippines

CAVITE STATE UNIVERSITY

Don Severino De Las Alas Campus
Indang, Cavite, Philippines

DIFFERENTIAL EQUATIONS: PROOF OF SOLUTION AND HOMOGENEOUS DE

Problem Set

Submitted to

Mr. Gin Karlo Villa

Instructor, DIET Faculty

In partial fulfillment

of the requirements for the degree

Bachelor of Science in Industrial Engineering

Submitted by:

Andam, Shenna Mae

Baybay, Alexa Gail

Colada, Sam Ibrahim

Domingo, Jarymon

Juanitas, Franz Adrielle

Mondares, Kevin

Pelorin. Kate Ritchel

Sangalang. Klein John Brent

Toledo, Paul Nikolai

PROOF OF SOLUTION
−2𝑥
1. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦''' − 3𝑦' + 2𝑦 = 0 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 = 𝑒

−2𝑥
𝑦 = 𝑒 𝑦''' − 3𝑦' + 2𝑦 = 0
−2𝑥 −2𝑥 −2𝑥 −2𝑥
𝑦' =− 2𝑒 (− 8𝑒 ) − 3(− 2𝑒 ) + 2(𝑒 )=0
−2𝑥 −2𝑥 −2𝑥 −2𝑥
𝑦'' = 4𝑒 − 8𝑒 + 6𝑒 + 2𝑒 =0
−2𝑥
𝑦''' =− 8𝑒 ⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

𝑑𝑥 1
2. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑑𝑦
=𝑥− 2
𝑦 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 = 2𝑥 − 4

𝑑𝑥 1
𝑦 = 2𝑥 − 4 𝑑𝑦
=𝑥− 2
𝑦
1
𝑦' = 2 2=𝑥− 2
(2𝑥 − 4)
2=𝑥−𝑥+2
⇒ 2 = 2, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

𝑡
3. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦'' − 𝑦 = 0 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑦1 = 𝑒 𝑎𝑛𝑑 𝑦2 = 𝑐𝑜𝑠ℎ 𝑡

𝑡
𝑦1 = 𝑒 𝑦'' − 𝑦 = 0
𝑡 𝑡 𝑡
𝑦1' = 𝑒 𝑒 −𝑒 = 0
𝑡
𝑦1'' = 𝑒 ⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

𝑦2 = 𝑐𝑜𝑠ℎ 𝑡 𝑦'' − 𝑦 = 0
𝑦2' = 𝑐𝑜𝑠ℎ 𝑡 𝑐𝑜𝑠ℎ 𝑡 − 𝑐𝑜𝑠ℎ 𝑡 = 0
𝑦2'' = 𝑐𝑜𝑠ℎ 𝑡 ⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

2 2
4. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡𝑦' − 𝑦 = 𝑡 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 = 3𝑡 + 𝑡

2 2
𝑦 = 3𝑡 + 𝑡 𝑡𝑦' − 𝑦 = 𝑡
2 2
𝑦' = 3 + 2𝑡 𝑡(3 + 2𝑡) − 3𝑡 + 𝑡 = 𝑡
2 2 2
3𝑡 + 2𝑡 − 3𝑡 + 𝑡 = 𝑡
2 2
⇒ 𝑡 = 𝑡 , 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
 −3𝑡 𝑡
5. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦'' + 2𝑦' − 3𝑦 = 0 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑦1 = 𝑒 𝑎𝑛𝑑 𝑦2 = 𝑒

−3𝑡
𝑦1 = 𝑒 𝑦'' + 2𝑦' − 3𝑦 = 0
−3𝑡 −3𝑡 −3𝑡 −3𝑡
𝑦1' =− 3𝑒 9𝑒 + 2(− 3𝑒 ) − 3(𝑒 )=0
−3𝑡 −3𝑡 −3𝑡 −3𝑡
𝑦1'' = 9𝑒 9𝑒 − 63𝑒 − 3𝑒 =0
⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

𝑡
𝑦2 = 𝑒 𝑦'' + 2𝑦' − 3𝑦 = 0
𝑡 𝑡 𝑡 𝑡
𝑦2' = 𝑒 𝑒 + 2𝑒 − 3𝑒 = 0
𝑡 𝑡 𝑡
𝑦2'' = 𝑒 3𝑒 − 3𝑒 = 0
⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

2𝑥
6. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦' = 2𝑦 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 = 𝑒

2𝑥
𝑦 =𝑒 𝑦' = 2𝑦
2𝑥 2𝑥 2𝑥
𝑦' = 2𝑒 2𝑒 = 2(𝑒 )
2𝑥 2𝑥
⇒ 2𝑒 = 2𝑒 , 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

𝑡 −𝑡 𝑡
7. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦'''' + 4𝑦''' + 3𝑦 = 𝑡 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑦1 = 3
𝑎𝑛𝑑 𝑦2 = 𝑒 + 3

𝑡
𝑦1 = 3
𝑦'''' + 4𝑦''' + 3𝑦 = 𝑡
1 𝑡
𝑦1' = 3
0 + 4(0) + 3( 3 ) = 𝑡
𝑦1'' = 0 ⇒ 𝑡 = 𝑡, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
𝑦1''' = 0
𝑦1'''' = 0

−𝑡 𝑡
𝑦2 = 𝑒 + 3
𝑦'''' + 4𝑦''' + 3𝑦 = 𝑡
−𝑡 1 −𝑡 −𝑡 −𝑡 𝑡
𝑦2' = − 𝑒 + 3
𝑒 + 4(− 𝑒 ) + 3(𝑒 + 3
)=𝑡
−𝑡 −𝑡 −𝑡 −𝑡
𝑦2'' = 𝑒 +0 𝑒 − 4𝑒 + 3𝑒 +𝑡=𝑡
−𝑡 −𝑡 −𝑡
𝑦2''' = − 𝑒 4𝑒 − 4𝑒 +𝑡=𝑡
−𝑡
𝑦'2''' = 𝑒 ⇒ 𝑡 = 𝑡, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
 −1
8. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑥𝑦' + 𝑦 = 2𝑥 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 = 𝑥 − 𝑥

−1
𝑦 = 𝑥 −𝑥 𝑥𝑦' + 𝑦 = 2𝑥
−2 −2 −1
𝑦' = 1 + 𝑥 𝑥(1 + 𝑥 ) + (𝑥 − 𝑥 ) = 2𝑥
−1 −1
𝑥 +𝑥 + 𝑥 − 𝑥 = 2𝑥
⇒ 2𝑥 = 2𝑥, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

−2𝑥 𝑥
9. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦''' − 3𝑦' + 2𝑦 = 0 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 3𝑒 + 4𝑒 .

−2𝑥 𝑥
𝑦 = 3𝑒 + 4𝑒
−2𝑥 𝑥
𝑦' =− 6𝑒 + 4𝑒
−2𝑥 𝑥
𝑦'' = 12𝑒 + 4𝑒
−2𝑥 𝑥
𝑦''' =− 24𝑒 + 4𝑒

𝑦''' − 3𝑦' + 2𝑦 = 0
−2𝑥 𝑥 −2𝑥 𝑥 −2𝑥 𝑥 𝑥
(− 24𝑒 + 4𝑒 ) − 3(− 6𝑒 + 4𝑒 ) + 2( 3𝑒 + 4𝑒 + 4𝑒 ) = 0
−2𝑥 𝑥 −2𝑥 𝑥 −2𝑥 𝑥
(− 24𝑒 + 4𝑒 ) + (18𝑒 − 12𝑒 ) + (6𝑒 + 8𝑒 ) = 0
−2𝑥 𝑥
(− 24 + 18 + 6)𝑒 + (4 − 12 − 8)𝑒 = 0
−2𝑥 𝑥
0𝑒 + 0𝑒 = 0
⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.

−𝑥 2𝑥
10. 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑦'' − 𝑦' − 2𝑦 = 0 𝑖𝑠 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 = 𝐶𝑒 1
+ 𝐶2𝑒

−𝑥 2𝑥
𝑦 = 𝐶1𝑒 + 𝐶2𝑒
2𝑥 −𝑥
𝑦' = 2𝐶1𝑒 − 𝐶2𝑒
2𝑥 −𝑥
𝑦'' = 4𝐶1𝑒 + 𝐶2𝑒

𝑦'' − 𝑦' − 2𝑦 = 0
2𝑥 −𝑥 2𝑥 −𝑥 −𝑥 2𝑥
4𝐶1𝑒 + 𝐶2𝑒 − 2𝐶1𝑒 − 𝐶2𝑒 − 2𝐶1𝑒 − 2𝑒 =0
2𝑥 −𝑥
(4 − 2 − 2)𝐶1𝑒 + (1 + 1 − 2)𝐶2𝑒 =0
⇒ 0 = 0, 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
HOMOGENEOUS DIFFERENTIAL EQUATION
2 2
1. (𝑥 + 𝑦 )𝑑𝑥 + 2𝑥𝑦 𝑑𝑦 = 0

2 2
(𝑥 + (𝑣𝑥) )𝑑𝑥 + 2𝑥(𝑣𝑥)(𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
2 2 2 2
(𝑥 + 𝑣 𝑥 )𝑑𝑥 + 2𝑥 𝑣(𝑣𝑑𝑥 + 𝑥𝑑𝑣) = 0
2 2 2 2 3
𝑥 (1 + 𝑣 )𝑑𝑥 + 2𝑥 𝑣 𝑑𝑥 + 2𝑥 𝑣𝑑𝑣 = 0
2 2 2 2
𝑥 [(1 + 𝑣 ) + 2𝑣 ]𝑑𝑥 + 2𝑥 𝑣𝑑𝑣 = 0
2 2 3 1
[𝑥 (1 + 3𝑣 )𝑑𝑥 + 2𝑥 𝑣𝑑𝑣 = 0] 2
𝑥
2 1
[(1 + 3𝑣 )𝑑𝑥 + 2𝑥𝑣𝑑𝑣 = 0] 2
𝑥(1+3𝑣 )
𝑑𝑥 2𝑣
𝑥
+ 2 = 0
1+3𝑣
𝑑𝑥 2𝑣
𝑥
=− 2
1+3𝑣

𝑑𝑢 2𝑣
∫ 𝑢
= ∫− 2 ; 𝑤ℎ𝑒𝑟𝑒 𝑢1 = 𝑥, 𝑎𝑛𝑑 𝑑𝑢1 = 𝑑𝑥
1+3𝑣

𝑑𝑢1 2𝑣 2
∫ 𝑢1
= ∫− 2 ; 𝑤ℎ𝑒𝑟𝑒 𝑢2 = 1 + 3𝑣 , 𝑎𝑛𝑑 𝑑𝑢2 = 6𝑣𝑑𝑣
1+3𝑣

−1 𝑑𝑢2
𝑙𝑛(𝑥) = ∫ 3
• 𝑢2

𝑙𝑛(𝑥) =−
1
3 | | 𝑙𝑛 𝑢2
2
𝑙𝑛|1 + 3𝑣 | + 𝐶
1
𝑙𝑛(𝑥) =− 3

1
2 −3
𝑙𝑛(𝑥) = 𝑙𝑛 1 + 3𝑣 | 1
| +𝐶
2 −3

𝑒
𝑙𝑛(𝑥)
=𝑒
𝑙𝑛 1+3𝑣 | | • 𝑒𝐶
1
2 −3
𝑥 = 1 + 3𝑣 | | • 𝐶
1
−3
| 𝑦 2|
𝑥 = |1 + 3( 𝑥 ) | • 𝐶
| |
2 1
𝑦 −3 1
[𝑥 = 𝐶 • (1 + 3 2 ) ] 2 −3
1
𝑥 (1+3
𝑦
)
2
𝑥

𝑥
⇒ 2 −3
1 =𝐶
𝑦
(1+3 2 )
𝑥
 2 2
2. 𝑥𝑦 𝑑𝑥 + (𝑥 + 𝑦 )𝑑𝑦

2 2 2
𝑣𝑦(𝑦)(𝑣𝑑𝑦 + 𝑦𝑑𝑣) + (𝑣 𝑦 + 𝑦 )𝑑𝑦 = 0
2 2 3 2 2 2
𝑣 𝑦 𝑑𝑦 + 𝑣𝑦 𝑑𝑣 + 𝑣 𝑦 𝑑𝑦 + 𝑦 𝑑𝑦 = 0
2 2 3 2 1
[2𝑣 𝑦 𝑑𝑦 + 𝑣𝑦 𝑑𝑣 + 𝑦 𝑑𝑦 = 0] 2
𝑦
2
2𝑣 𝑑𝑦 + 𝑣𝑦𝑑𝑣 + 𝑑𝑦 = 0
2 1
[(2𝑣 + 1)𝑑𝑦 + 𝑣𝑦𝑑𝑣 = 0] 2
𝑦(2𝑣 +1)
𝑑𝑦 𝑣𝑑𝑣
𝑦
+ 2 = 0
2𝑣 +1

𝑑𝑦 𝑣𝑑𝑣
∫ 𝑦
+∫ 2 = ∫ 0; 𝑤ℎ𝑒𝑟𝑒 𝑢1 = 𝑦, 𝑎𝑛𝑑 𝑑𝑢1 = 𝑑𝑦
2𝑣 +1

𝑑𝑢1 𝑣𝑑𝑣 2 𝑑𝑢2

∫ 𝑢1
+∫ 2 = 𝐶; 𝑤ℎ𝑒𝑟𝑒 𝑢2 = 2𝑣 + 1, 𝑑𝑢2 = 𝑣𝑑𝑣, 𝑎𝑛𝑑 4
= 𝑣𝑑𝑣
2𝑣 +1
𝑑𝑢2
4
𝑙𝑛(𝑦) + ∫ 𝑢2
= 𝐶

1 𝑑𝑢2
𝑙𝑛(𝑦) + 4
∫ 𝑢2
=𝐶
1
𝑙𝑛(𝑦) + 4
𝑙𝑛(𝑢2) = 𝐶
1 2
[𝑙𝑛(𝑦) + 4
𝑙𝑛(2𝑣 + 1) = 𝐶] • 4
2
4𝑙𝑛(𝑦) + 𝑙𝑛(2𝑣 + 1) = 4𝐶
4 2
𝑙𝑛(𝑦) + 𝑙𝑛(2𝑣 + 1) = 𝐶
4 2
𝑙𝑛(𝑦) 𝑙𝑛(2𝑣 +1)
𝑒 •𝑒 = 𝐶
4 2
𝑦 • (2𝑣 + 1) = 𝐶
4 𝑥 2
𝑦 • [2( 𝑦 ) + 1] = 𝐶
2
4 𝑥
𝑦 • [2( 2 ) + 1] = 𝐶
𝑦
2 2
4 𝑥 +𝑦
𝑦 • [2( 2 )] = 𝐶
𝑦

2 2 2
⇒ 𝑦 (2𝑥 + 𝑦 ) = 𝐶
 2
3. 𝑣 𝑑𝑥 + 𝑥(𝑥 + 𝑣)𝑑𝑣 = 0; 𝑙𝑒𝑡 𝑥 = 𝑢𝑣, 𝑎𝑛𝑑 𝑑𝑥 = 𝑢𝑑𝑣 + 𝑣𝑑𝑢

2
𝑣 (𝑢𝑑𝑣 + 𝑣𝑑𝑢) + 𝑢𝑣(𝑢𝑣 + 𝑣)𝑑𝑣 = 0
2 3 2 2 2
𝑢𝑣 𝑑𝑣 + 𝑣 𝑑𝑢 + 𝑢 𝑣 𝑑𝑣 + 𝑢𝑣 𝑑𝑣 = 0
2 2 2 3
2𝑢𝑣 𝑑𝑣 + 𝑢 𝑣 𝑑𝑣 + 𝑣 𝑑𝑢 = 0
2 3 1
[𝑢𝑣 (2 + 𝑢)𝑑𝑣 + 𝑣 𝑑𝑢 = 0] 3
𝑣 𝑢(2+𝑢)
𝑑𝑣 𝑑𝑢
𝑣
+ 𝑢(2+𝑢)
= 0

1 𝐴 𝐵
( 𝑢(2+𝑢) = 𝑢
+ 2+𝑢
) 𝑢 (2 + 𝑢)
1 1
1 = 𝐴(2 + 𝑢) + 𝐵(𝑢); 𝑤ℎ𝑒𝑟𝑒 𝑢𝑎 = 0, 1𝑎 = 𝐴(2), 𝐴 = 2
, 𝑎𝑛𝑑 𝑢𝑏 =− 2, 1𝑏 = 𝐵(− 2), 𝐵 =− 2
1 1/2 1/2
𝑢(2+𝑢)
= 𝑢
− 2+𝑢

𝑑𝑣 1/2 1/2
∫ 𝑣
+ ∫( 𝑢
− 2+𝑢
)𝑑𝑢 = ∫ 0; 𝑤ℎ𝑒𝑟𝑒 𝑢𝐶 = 𝑣, 𝑎𝑛𝑑 𝑑𝑢𝐶 = 𝑑𝑣

𝑑𝑢𝑐 1 𝑑𝑢𝐴 𝑑𝑢𝐵

∫ 𝑢𝐶
+ 2
∫( 𝑢
− 2+𝑢
)𝑑𝑢 = 𝐶
1 1
[𝑙𝑛(𝑣) + 2
𝑙𝑛(𝑢) − 2
𝑙𝑛(2 + 𝑢) = 𝐶] • 2
2𝑙𝑛(𝑣) + 𝑙𝑛(𝑢) − 𝑙𝑛(2 + 𝑢) = 𝐶
2
𝑙𝑛(𝑣) + 𝑙𝑛(𝑢) − 𝑙𝑛(2 + 𝑢) = 𝐶
| 𝑢𝑣2 |
𝑙𝑛 | 2+𝑢 | = 𝐶
| |
| 𝑢𝑣2 |
𝑙𝑛 | 2+𝑢 |
| |
𝑒 = 𝐶
2
𝑢𝑣
2+𝑢
+ 𝐶
2 𝑥
𝑢𝑣 = 𝐶 (2 + 𝑢); 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑢 = 𝑣
𝑥 2 𝑥
𝑣
𝑣 = 𝐶 (2 + 𝑣
)
𝑥 2 2𝑣+𝑥
𝑣
𝑣 = 𝐶( 𝑣
)
2 1
[𝑥𝑣 = 𝐶 (2𝑣 + 𝑥)] 2𝑣+𝑥

2
𝑥𝑣
⇒ 2𝑣+𝑥
=𝐶
 2 2
4. 2𝑥𝑣𝑦' = 𝑥 + 𝑦
2 2 1
[2𝑥𝑣𝑦' = 𝑥 + 𝑦 ] 2𝑥𝑦
2 2
𝑥 +𝑦
𝑦' = 2𝑥𝑦
𝑥 𝑦
𝑦' = 2𝑦
+ 2𝑥
𝑦
1 𝑥
𝑦' = 2𝑦 + 2
; 𝑤ℎ𝑒𝑟𝑒 𝑣 = 𝑦/𝑥, 𝑎𝑛𝑑 𝑦' = 𝑥𝑣' + 𝑣
𝑥
1 𝑣
𝑥𝑣' + 𝑣 = 2𝑣
+ 2
1 𝑣
𝑥𝑣' = 2𝑣
− 2
2
𝑣 −1
𝑣' = 2𝑣𝑥
2𝑣 1
2 𝑣' =− 𝑥
𝑣 −1

2𝑣 1 2
∫ 2 𝑑𝑣 =− ∫ 𝑥
𝑑𝑥; 𝑤ℎ𝑒𝑟𝑒 𝑢1 = 𝑣 − 1, 𝑑𝑢1 = 2𝑣, 𝑎𝑛𝑑 𝑢2 = 𝑥, 𝑑𝑢2 = 𝑑𝑥
𝑣 −1

𝑑𝑢1 𝑑𝑢2
∫ 𝑢1
=− ∫ 𝑢2

|2 |
𝑙𝑛 𝑣 − 1 =− 𝑙𝑛|𝑥| + 𝐶
2 −1
𝑙𝑛|𝑣 − 1| = 𝑙𝑛|𝑥| + 𝐶
2

𝑒
|
𝑙𝑛 𝑣 −1 | = 𝑒𝑙𝑛|𝑥|−1 • 𝑒𝐶
2 −1
𝑣 − 1 =𝑥 • 𝐶
2 −1
𝑣 = 1 +𝑥 • 𝐶
2 1
𝑣 = 1+ 𝑥
• 𝐶
2 𝐶
𝑣 = 1+ 𝑥
𝑦 2 𝐶
(𝑥) = 1 + 𝑥
2
𝑦 𝐶 2
[ 2 = 1+ 𝑥
]• 𝑥
𝑥
2 2
𝑦 = 𝑥 + 𝐶𝑥
2 2 1
[𝑦 − 𝑥 = 𝐶𝑥] 𝑥

2 2
𝑦 −𝑥
⇒ 𝑥
=𝐶
 2 2
5. (𝑥 + 𝑦 )𝑑𝑥 + 𝑥𝑦 𝑑𝑦 = 0; ; 𝑒𝑡 𝑦 = 𝑥𝑣, 𝑎𝑛𝑑 𝑑𝑦 = 𝑥𝑑𝑣 + 𝑣𝑑𝑥
2 2 2 2
(𝑥 + 𝑥 𝑣 )𝑑𝑥 + 𝑥 𝑣(𝑥𝑑𝑣 + 𝑣𝑑𝑥) = 0
2 2 2 3 2 2
(𝑥 + 𝑥 𝑣 )𝑑𝑥 + 𝑥 𝑣𝑑𝑣 + 𝑥 𝑣 𝑑𝑥 = 0
2 2 2 3 1
[𝑥 + 2𝑥 𝑣 𝑑𝑥 + 𝑥 𝑣𝑑𝑣 = 0] 2
𝑥
2 1
[(1 − 2𝑥 )𝑑𝑥 + 𝑥𝑣𝑑𝑣 = 0] 2
𝑥(1+2𝑣 )
𝑑𝑥 𝑣𝑑𝑣
𝑥
+ 2 = 0
1+2𝑣

𝑑𝑥 𝑣𝑑𝑣 2 𝑑𝑢2
∫ 𝑥
+∫ 2 = ∫ 0; 𝑤ℎ𝑒𝑟𝑒 𝑢1 = 𝑥, 𝑑𝑢1 = 𝑑𝑥, 𝑎𝑛𝑑 𝑢2 = 1 + 2𝑣 , 𝑑𝑢2 = 4𝑣𝑑𝑣 𝑜𝑟 4
= 𝑣𝑑𝑣
1+2𝑣
𝑑𝑢2
𝑑𝑢1 4
∫ 𝑢1
+∫ 𝑢2
= 𝐶

1 𝑑𝑢2
𝑙𝑛(𝑥) + 4
∫ 𝑢2
= 𝐶

1 2
𝑙𝑛(𝑥) + 4
𝑙𝑛(1 + 2𝑣 ) = 𝐶
1 2
[𝑙𝑛(𝑥) + 4
𝑙𝑛(1 + 2𝑣 ) = 𝐶] • 4
2
4𝑙𝑛(𝑥) + 𝑙𝑛(1 + 2𝑣 ) = 𝐶
4 2
𝑙𝑛(𝑥) + 𝑙𝑛(1 + 2𝑣 ) = 𝐶
4 2
𝑙𝑛(𝑥) 𝑙𝑛(1+2𝑣 )
𝑒 •𝑒 = 𝐶
4 2 𝑦
𝑥 • (1 + 2𝑣 ) = 𝐶; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑣 = 𝑥
4 𝑦 2
𝑥 • (1 + 2( 𝑥 ) ) = 𝐶
2
4 𝑦
𝑥 • (1 + 2 2 ) =𝐶
𝑥
2 2
4 𝑥 +2𝑦
𝑥 • ( 2 ) = 𝐶
𝑥

2 2 2
⇒ 𝑥 (𝑥 + 2𝑦 ) = 𝐶
 𝑦 𝑦 𝑦 𝑦
6. (𝑥 𝑠𝑖𝑛 𝑥
− 𝑦 𝑐𝑜𝑠 𝑥 )𝑑𝑥 + 𝑥 𝑐𝑜𝑠 𝑥
𝑑𝑦 = 0; 𝑙𝑒𝑡 𝑦 = 𝑣𝑥, 𝑑𝑦 = 𝑥𝑑𝑣 + 𝑣𝑑𝑥, 𝑎𝑛𝑑 𝑣 = 𝑥
(𝑥 𝑠𝑖𝑛 𝑣 − 𝑥𝑣 𝑐𝑜𝑠 𝑣)𝑑𝑥 + 𝑥 𝑐𝑜𝑠 𝑣(𝑥𝑑𝑣 + 𝑣𝑑𝑥) = 0
2 2 2 1
[𝑥 𝑠𝑖𝑛 𝑣 𝑑𝑥 − 𝑥𝑣 𝑐𝑜𝑠 𝑑𝑥 + 𝑥 𝑐𝑜𝑠 𝑣 𝑑𝑣 + 𝑥 𝑐𝑜𝑠 𝑣 𝑑𝑥 = 0] 𝑥
2 2
𝑠𝑖𝑛 𝑣 𝑑𝑥 − 𝑣 𝑐𝑜𝑠 𝑑𝑥 + 𝑥𝑣 𝑐𝑜𝑠 𝑑𝑥 + 𝑥 𝑐𝑜𝑠 𝑣 𝑑𝑣 = 0
1
[𝑠𝑖𝑛 𝑣 𝑑𝑥 + 𝑥 𝑐𝑜𝑠 𝑣 𝑑𝑣 = 0] 𝑥(𝑠𝑖𝑛 𝑣)
𝑑𝑥 𝑐𝑜𝑠 𝑣 𝑐𝑜𝑠 𝑣
𝑥
+ 𝑠𝑖𝑛 𝑣
𝑑𝑣 = 0; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑐𝑜𝑡 𝑣 = 𝑠𝑖𝑛 𝑣

𝑑𝑥
𝑥
+ 𝑐𝑜𝑡 𝑣 𝑑𝑣 = 0

𝑑𝑥
∫ 𝑥
+ ∫ 𝑐𝑜𝑡 𝑣 𝑑𝑣 = ∫ 0; 𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑥, 𝑎𝑛𝑑 𝑑𝑢 = 𝑑𝑥

𝑑𝑢
∫ 𝑢
+ 𝑙𝑛|𝑠𝑖𝑛 𝑣| = 𝐶

𝑙𝑛|𝑥| + 𝑙𝑛|𝑠𝑖𝑛 𝑣| = 𝐶
𝑙𝑛|𝑥| 𝑙𝑛|𝑠𝑖𝑛 𝑣|
𝑒 •𝑒 = 𝐶
𝑦
𝑥 • 𝑠𝑖𝑛 𝑣 = 𝐶; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑣 = 𝑥
𝑦
𝑥 • 𝑠𝑖𝑛 𝑥
= 𝐶

𝑦
⇒ 𝑥𝑠𝑖𝑛( 𝑥 ) = 𝐶
7. 𝑥𝑦' = 2𝑥 + 3𝑦
3𝑦 𝑑𝑦
𝑦' = 2 + 𝑥
; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑦' = 𝑣𝑥 𝑖𝑠 𝑑𝑥
𝑑𝑣 3𝑣𝑥
𝑑𝑥
𝑥+ 𝑣 =2 + 𝑥
𝑑𝑣
𝑑𝑥
𝑥 + 𝑣 = 2 + 3𝑣
𝑑𝑣
𝑑𝑥
𝑥 = 2 + 3𝑣 − 𝑣
𝑑𝑣 𝑑𝑥
[ 𝑑𝑥 𝑥 = 2 + 2𝑣] 2+2𝑣
𝑑𝑣 𝑑𝑥
2+2𝑣
= 𝑥

𝑑𝑣 𝑑𝑥
∫ 2+2𝑣
=∫ 𝑥

1 𝑑𝑣 𝑑𝑥
2
∫ 1+2𝑣
=∫ 𝑥
; 𝑤ℎ𝑒𝑟𝑒 𝑢1 = 1 + 𝑣, 𝑑𝑢 = 𝑑𝑣, 𝑎𝑛𝑑 𝑢2 = 𝑥, 𝑑𝑢2 = 𝑥

1 𝑑𝑢1 𝑑𝑢2
2
∫ 𝑢1
=∫ 𝑢2

1
2
𝑙𝑛|1 + 𝑣| = 𝑙𝑛|𝑥| + 𝐶
1
[ 2 𝑙𝑛|1 + 𝑣| = 𝑙𝑛|𝑥| + 𝐶] •2
𝑙𝑛|1 + 𝑣| = 2𝑙𝑛|𝑥| + 𝐶
2
𝑙𝑛|1 + 𝑣| = 𝑙𝑛|𝑥| + 𝐶
2
𝑙𝑛|1+𝑣| 𝑙𝑛|𝑥| 𝐶
𝑒 =𝑒 •𝑒
2 𝑦
(1 + 𝑣) = 𝑥 • 𝐶; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑣 = 𝑥
𝑦 2
(1 + 𝑥
) =𝑥 • 𝐶
𝑥+𝑦 2
( 𝑥
) =𝑥 • 𝐶
1
[𝑥 + 𝑦 = 𝐶𝑥] 𝑥

𝑥+𝑦
⇒ 𝑥
=𝐶
 2 2
8. 𝑥𝑑𝑦 = (𝑦 + 𝑥 + 𝑦 )𝑑𝑥
𝑑𝑦 2 2 𝑑𝑦 𝑑𝑣
𝑥 𝑑𝑥
=𝑦+ 𝑥 + 𝑦 ; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑦 = 𝑣𝑥 𝑖𝑠 𝑑𝑥
= 𝑑𝑥
𝑥+𝑣
𝑑𝑣 2 2
𝑥( 𝑑𝑥 𝑥 + 𝑣) = 𝑦 + 𝑥 +𝑦
𝑑𝑣 2 2
𝑥( 𝑑𝑥 𝑥 + 𝑣 = 𝑣𝑥 + 𝑥 + (𝑣𝑥)

𝑑𝑣 2 2 2
𝑥( 𝑑𝑥 𝑥 + 𝑣 = 𝑣𝑥 + 𝑥 + (𝑣 𝑥 )
𝑑𝑣 2 2
𝑥( 𝑑𝑥 𝑥 + 𝑣 = 𝑣𝑥 + 𝑥 + (1 + 𝑣 )
𝑑𝑣 2 1
[𝑥( 𝑑𝑥 𝑥 + 𝑣 = 𝑣𝑥 + 𝑥 1 + 𝑣 ] 𝑥

𝑑𝑣 2
𝑑𝑥
𝑥 + 𝑣 =𝑣 + 1 +𝑣
𝑑𝑣 2
𝑑𝑥
𝑥 = 𝑣 −𝑣 + 1 +𝑣
𝑑𝑣 2 𝑑𝑥
[ 𝑑𝑥 𝑥 = 1 +𝑣 ] 2
𝑥( 1+𝑣 )
𝑑𝑣 𝑑𝑥
2
= 𝑥
1+𝑣

𝑑𝑣 𝑑𝑥 𝑑𝑢 | 2 2|
∫ =∫ 𝑥
; 𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 = 𝑙𝑛|𝑢 + 𝑢 + 𝑎 |+ 𝐶
1+𝑣
2 2
𝑢 +𝑎
2 | |
| 2| 𝑑𝑥
𝑙𝑛|1 + 1 + 𝑣 |= ∫ 𝑥
; 𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑥, 𝑑𝑢 = 𝑑𝑥
| |
| 2| 𝑑𝑢
𝑙𝑛|1 + 1 + 𝑣 |= ∫ 𝑢
| |
| 2|
𝑙𝑛|1 + 1 + 𝑣 | = 𝑙𝑛|𝑥| + 𝐶
| |
| 2|
𝑙𝑛|1+ 1+𝑣 |
| | 𝑙𝑛|𝑥| 𝐶
𝑒 =𝑒 •𝑒
2
1+ 1 +𝑣 = 𝑥 • 𝐶
𝑦 2
1+ 1 + ( 𝑥 ) = 𝐶𝑥
2
𝑦
1+ 1 + 2 = 𝐶𝑥
𝑥
2 2
𝑥 + 𝑥 +𝑦
𝑥
= 𝐶𝑥
2 2
⇒𝑥 + 𝑥 +𝑦 = 𝐶
 2 2
9. (𝑥 − 3𝑦 )𝑑𝑥 − 2𝑥𝑦 𝑑𝑦 = 0; 𝑙𝑒𝑡 𝑦 = 𝑢𝑥, 𝑎𝑛𝑑 𝑑𝑦 = 𝑢𝑑𝑥 + 𝑥𝑑𝑢
2 2
(𝑥 − 3𝑦 )𝑑𝑥 − 2𝑥𝑦 𝑑𝑦 = 0
2 2
(𝑥 + 3(𝑢𝑥) )𝑑𝑥 − 2𝑥(𝑢𝑥)(𝑢𝑑𝑥 + 𝑥𝑑𝑢)
2 2 2 2 2 3
𝑥 𝑑𝑥 + 3𝑢 𝑥 𝑑𝑥 − 2𝑥 𝑢 𝑑𝑥 2𝑥 𝑢𝑑𝑢 = 0
2 2 2 3
𝑥 𝑑𝑥 + 𝑢 𝑥 𝑑𝑥 − 2𝑥 𝑢𝑑𝑢 = 0
2 2 3
𝑥 + (1 + 𝑢 )𝑑𝑥 = 2𝑥 𝑢𝑑𝑢
1 2𝑢 2
𝑥
𝑑𝑥 = 2 𝑑𝑢; 𝑤ℎ𝑒𝑟𝑒 𝑢 = 1 + 𝑢 , 𝑎𝑛𝑑 𝑑𝑢 = 2𝑢𝑑𝑢
1+𝑢

1 𝑑𝑢
∫ 𝑥
𝑑𝑢 = ∫ 𝑢

𝑙𝑛 |𝑥| = 𝑙𝑛|𝑢| + 𝐶
2
𝑙𝑛 |𝑥| = 𝑙𝑛 1 + 𝑢 + 𝐶 | |
𝑦 2
𝑙𝑛 |𝑥| (𝑙𝑛||1+( 𝑥 ) ||+𝐶)
𝑒 =𝑒
𝑦 2 𝐶
𝑥 = ||1 + ( 𝑥 ) || • 𝑒
2 2
2 𝑥 +𝑦 2
𝑥 [𝑥 = 𝐶 • ( 2 )]𝑥
𝑥
3 2 2 1
[𝑥 = 𝐶(𝑥 + 𝑦 )] 2 2
𝑥 +𝑦

3
𝑥
⇒ 2 2 =𝐶
𝑥 +𝑦
10. (𝑥 − 𝑦)𝑑𝑥 + 𝑥 𝑑𝑦 = 0; 𝑙𝑒𝑡 𝑦 = 𝑢𝑥, 𝑎𝑛𝑑 𝑑𝑦 = 𝑢𝑑𝑥 + 𝑥𝑑𝑢
(𝑥 − 𝑦)𝑑𝑥 + 𝑥 𝑑𝑦 = 0
(𝑥 − 𝑢𝑥)𝑑𝑥 + 𝑥 (𝑢𝑑𝑥 + 𝑥𝑑𝑢) = 0
2
𝑥𝑑𝑥 − 𝑢𝑥 𝑑𝑥 + 𝑥𝑢 𝑑𝑥 + 𝑥 𝑑𝑦 = 0
2
𝑥𝑑𝑥 𝑥 𝑑𝑢
2 =− 2
𝑥 𝑥

1
∫ 𝑥
𝑑𝑥 = ∫− 𝑑𝑢

𝑙𝑛|𝑥| =− 𝑢 + 𝐶
𝑦
𝑙𝑛|𝑥| =− 𝑥
+𝐶
𝑙𝑛|𝑥|
𝑒 =− 𝑢 + 𝐶
𝑥 =− 𝑢 + 𝐶
𝑦
𝑥 =− 𝑥
+𝐶
𝑦
𝑥− 𝑥
=𝐶

2
𝑥 −𝑦
⇒ 𝑥
=𝐶
REFERENCES

Al-johani, A. S. Ordinary differential equations. Magnum Publishing, N,Y. c.2016

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