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Solution 2.2

The document contains examples of solving differential equations using integration techniques. Many of the examples involve taking the integral of one side of a differential equation and setting it equal to the integral of the other side. The solutions for y(x) are then obtained by taking the inverse of the integral.

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anastasia chkhenkeli
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39 views4 pages

Solution 2.2

The document contains examples of solving differential equations using integration techniques. Many of the examples involve taking the integral of one side of a differential equation and setting it equal to the integral of the other side. The solutions for y(x) are then obtained by taking the inverse of the integral.

Uploaded by

anastasia chkhenkeli
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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𝑑𝑦

2. = (𝑥 + 1)2
𝑑𝑥

𝑑𝑦 = (𝑥 + 1)2 𝑑𝑥

∫ 𝑑𝑦 = ∫(𝑥 + 1)2 𝑑𝑥

(𝑥 + 1)3
𝑦= +𝑐
3
4. 𝑑𝑦 − (𝑦 − 1)2 𝑑𝑥 = 0

𝑑𝑦 = (𝑦 − 1)2 𝑑𝑥
1
∫ 𝑑𝑦 = ∫ 𝑑𝑥
(𝑦 − 1)2

𝑦−1=𝑚 𝑑𝑦 = 𝑑𝑚
1
∫ 𝑑𝑦 = ∫ 𝑑𝑥
𝑚2
1 1
−𝑚−1 = 𝑥 + 𝑐 − =𝑥+𝑐 𝑦 =1−
𝑦−1 𝑥+𝑐
𝑑𝑦
6. + 2𝑥𝑦 2 = 0
𝑑𝑥

1
∫ 𝑑𝑦 = ∫ −2𝑥 𝑑𝑥
𝑦2
𝑥2 1
−𝑦 −1 = −2 +𝑐 𝑦=
2 𝑥2 + 𝑐1
𝑑𝑦
8.𝑒 𝑥 𝑦 𝑑𝑥 = 𝑒 −𝑦 + 𝑒 −2𝑥−𝑦

𝑑𝑦
𝑒𝑥𝑦 = 𝑒 −𝑦 + 𝑒 −2𝑥 𝑒 −𝑦
𝑑𝑥
𝑑𝑦
𝑒𝑦𝑒 𝑥𝑦 = 𝑒 𝑦 𝑒 −𝑦 + 𝑒 −2𝑥 𝑒 −𝑦 𝑒 𝑦
𝑑𝑥
1 + 𝑒 −2𝑥
𝑒 𝑦 𝑦 𝑑𝑦 = 𝑑𝑥 ∫ 𝑒 𝑦 𝑦 𝑑𝑦 = ∫(𝑒 −𝑥 + 𝑒 −3𝑥 )𝑑𝑥
𝑒𝑥

∫ 𝑒 𝑦 𝑦 𝑑𝑦 = 𝑦𝑒 𝑦 − ∫ 𝑒 𝑦 𝑑𝑦 = 𝑦𝑒 𝑦 − 𝑒 𝑦 = 𝑒 𝑦 (𝑦 − 1) + 𝑐
𝑒 −3𝑥
𝑒 𝑦 (𝑦 − 1) = −𝑒 −𝑥 − +𝑐
3
𝑑𝑦 2𝑦+3
10. 𝑑𝑥 = (4𝑥+5)2

𝑑𝑦 𝑑𝑥
∫ 2
=∫
(2𝑦 + 3) (4𝑥 + 5)2

2𝑦 + 3 = 𝑚 2𝑑𝑦 = 𝑑𝑚
4𝑥 + 5 = 𝑛 4𝑑𝑥 = 𝑑𝑛
1 1
∫ 𝑚−2 𝑑𝑚 = ∫ 𝑛−2 𝑑𝑛
2 4
1 1
− 𝑚−1 = − 𝑛−1 + 𝑐
2 4
2 1
= +𝑐
(2𝑦 + 3) (4𝑥 + 5)
12. sin 3𝑥 𝑑𝑥 + 2𝑦 cos 3 3𝑥𝑑𝑦 = 0
− sin 3𝑥
∫ 2𝑦 𝑑𝑦 = ∫ 𝑑𝑥
cos 3 3𝑥
𝑦2
2 = − ∫ tan 3𝑥 sec 2 3𝑥 𝑑𝑥
2
sec 3𝑥 = 𝑚 3𝑠𝑒𝑐3𝑥 𝑡𝑎𝑛3𝑥 𝑑𝑥 = 𝑑𝑚
1 1 𝑚2 1
− ∫ tan 3𝑥 𝑠𝑒𝑐3𝑥 𝑠𝑒𝑐3𝑥 𝑑𝑥 = − ∫ 𝑚𝑑𝑚 = − + 𝑐 = − sec 2 3𝑥 + 𝑐
3 3 2 6
1
𝑦 2 = − sec 2 3𝑥 + 𝑐
6
14.𝑥(1 + 𝑦 2 )1/2 𝑑𝑥 = 𝑦(1 + 𝑥 2 )1/2 𝑑𝑦
𝑦 𝑥
∫ 2 1/2
𝑑𝑦 = ∫ 𝑑𝑥
(1 + 𝑦 ) (1 + 𝑥 2 )1/2

1 + 𝑦2 = 𝑛 2𝑦𝑑𝑦 = 𝑑𝑛

1 + 𝑥2 = 𝑚 2𝑥𝑑𝑥 = 𝑑𝑚
1 1
∫ 𝑛−1/2 𝑑𝑛 = ∫ 𝑚−1/2 𝑑𝑚 2𝑛1/2 = 2𝑚1/2 + 𝑐
2 2
(1 + 𝑦 2 )1/2 = (1 + 𝑥 2 )1/2 + 𝑐1
𝑑𝑄
16. 𝑑𝑡 = 𝑘(𝑄 − 70)
𝑑𝑄
∫ = ∫ 𝑘𝑑𝑡
𝑄 − 70
𝑙𝑛|𝑄 − 70| = 𝑘𝑡 + 𝑐

𝑄 − 70 = 𝑒 𝑘𝑡+𝑐

𝑄 = 𝑐1 𝑒 𝑘𝑡 + 70
𝑑𝑁
18. 𝑑𝑡
+ 𝑁 = 𝑁𝑡𝑒 𝑡+2

𝑑𝑁
= 𝑁(𝑡𝑒 𝑡+2 + 1)
𝑑𝑡
1
∫ 𝑑𝑁 = ∫(𝑡𝑒 𝑡+2 + 1)𝑑𝑡
𝑁

𝑙𝑛|𝑁| = 𝑡𝑒 𝑡+2 − 𝑒 𝑡+2 − 𝑡 + 𝑐


𝑑𝑦 𝑥𝑦+2𝑦−𝑥−2
20. 𝑑𝑥 = 𝑥𝑦−3𝑦+𝑥−3

𝑥𝑦 + 2𝑦 − 𝑥 − 2 (𝑦 − 1)(𝑥 + 2)
=
𝑥𝑦 − 3𝑦 + 𝑥 − 3 (𝑦 + 1)(𝑥 − 3)
𝑦+1 𝑥+2
∫ 𝑑𝑦 = ∫ 𝑑𝑥
𝑦−1 𝑥−3
2 5
∫1+ 𝑑𝑦 = ∫ 1 + 𝑑𝑥
𝑦−1 𝑥−3
𝑦 + 2𝑙𝑛|𝑦 − 1| = 𝑥 + 5𝑙𝑛|𝑥 − 3| + 𝑐

𝑒 𝑦 (𝑦 − 1)2 = 𝑒 𝑥 (𝑥 − 3)5 𝑐1
(𝑦 − 1)2
= 𝑐1 𝑒 𝑥−𝑦
(𝑥 − 3)5
𝑑𝑦
22. (𝑒 𝑥 + 𝑒 −𝑥 ) 𝑑𝑥 = 𝑦 2

1 1
∫ 2
𝑑𝑦 = ∫ 𝑥 𝑑𝑥
𝑦 𝑒 + 𝑒 −𝑥
1 1
− =∫ 𝑥 𝑑𝑥 𝑒𝑥 = 𝑚 𝑒 𝑥 𝑑𝑥 = 𝑑𝑚
𝑦 𝑒 + 𝑒 −𝑥
1 𝑑𝑚
∫ 𝑑𝑥 = ∫ = tan−1 𝑚 + 𝑐 = tan−1 𝑒 𝑥 + 𝑐
𝑒 𝑥 + 𝑒 −𝑥 1 + 𝑚2
1
𝑦 = − tan−1 𝑒 𝑥 +𝑐
𝑑𝑦 5
24. + 2𝑦 = 1 𝑦(0) =
𝑑𝑡 2

𝑑𝑦
∫ = ∫ 𝑑𝑡
1 − 2𝑦
1
− 𝑙𝑛|1 − 2𝑦| = 𝑡 + 𝑐
2
5
1 − 2𝑦 = 𝑒 −2𝑡 𝑐1 1−2 = 𝑐1 𝑐1 = 4
2
1
𝑦= − 2𝑒 −2𝑡
2
𝑑𝑦 1
30. 𝑑𝑥 = 𝑦 2 sin2 𝑥 𝑦(−2) = 3
𝑥 𝑥
1 𝑑𝑦 𝑑𝑥
∫ 2
= ∫ sin2 𝑥
−2 𝑦 𝑑𝑡 −2 𝑑𝑡

𝑥
𝑥 𝑑𝑥
−𝑦(𝑡)−1 | = ∫ sin2 𝑥
−2 −2 𝑑𝑡
𝑥
𝑑𝑥
−𝑦(𝑥)−1 + −𝑦(2)−1 = ∫ sin2 𝑥
−2 𝑑𝑡
𝑥
𝑑𝑥
−𝑦(𝑥)−1 = −𝑦(2)−1 + ∫ sin2 𝑥
−2 𝑑𝑡
𝑥
𝑑𝑥
−𝑦(𝑥)−1 = 3 − ∫ sin2 𝑥
−2 𝑑𝑡
1
𝑦=
𝑥 𝑑𝑥
3 − ∫−2 sin2 𝑥
𝑑𝑡

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