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Elimination of Arbitratry Constants
Solve the following:
2

1
...
3𝑥 − 𝑥𝑦 = 𝐶
Solution:
2

6𝑥 − (2𝑥𝑦𝑦' + 𝑦 ) = 0
2

6𝑥 − 2𝑥𝑦𝑦' − 𝑦 = 0
6𝑥 − 2𝑥𝑦

𝑑𝑦
𝑑𝑥

2

−𝑦 = 0
2

6𝑥𝑑𝑥 − 2𝑥𝑑𝑦' − 𝑦 𝑑𝑥 = 0
2

(6𝑥 − 𝑦 )𝑑𝑥 − 2𝑥𝑦𝑑𝑦 = 0 Ans

2

3
...
𝑦 = 𝐶1 + 𝐶2𝑒
Solution:
3𝑥

𝑦' = 3𝐶2𝑒 (Eq
...
2)
Multiply (Eq
...
2);
3𝑦' = 9𝐶2𝑒

3𝑥

3𝑥

𝑦'' = 9𝐶2𝑒

___________
3𝑦' − 𝑦'' = 0
Thus,
𝑦'' − 3𝑦' = 0 Ans

𝑥

−𝑥

5
...
𝑦 = 𝑥 + 𝐶1𝑒 + 𝐶2𝑒

−𝑥

Solution:
𝑥

−𝑥

𝑦' = 1 + 𝐶1𝑒 − 𝐶2𝑒
𝑥

−𝑥

𝑦'' = 𝐶1𝑒 + 𝐶2𝑒

𝑥

But 𝑦 − 𝑥 = 𝐶1𝑒 + 𝐶2𝑒
Thus,
𝑦'' = 𝑦 − 𝑥
𝑦'' − 𝑦 =− 𝑥 Ans

−𝑥

2𝑥

3𝑥

7
...
1)
2𝑥

𝑦' = 2𝐶1𝑒

3𝑥

+ 3𝐶2𝑒 (Eq
...
3)

Multiply (Eq
...
2);
2𝑥

3𝑥

2𝑦 = 2𝐶1𝑒

2𝑥

𝑦' = 2𝐶1𝑒

+ 2𝐶2𝑒

3𝑥

+ 3𝐶2𝑒

____________________
3𝑥

2𝑦' − 𝑦'' =− 𝐶2𝑒 (Eq
...
2) by 2 then subtract (Eq
...
5)
Multiply (Eq
...
5);
3𝑥

6𝑦' − 3𝑦'' =− 3𝐶2𝑒

3𝑥

2𝑦' − 𝑦'' =− 3𝐶2𝑒

___________________
6𝑦 − 3𝑦' − 2𝑦' + 𝑦'' = 0
Thus,
𝑦'' − 5𝑦' + 6𝑦 = 0 Ans

3𝑥

8
...
𝑦 = 𝐶1 + 𝐶2𝑒
Solution:

−4𝑥

𝑦' =− 4𝐶2𝑒

−4𝑥

𝑦'' = 16𝐶2𝑒

(Eq
...
2)

Multiply (Eq
...
2);
−4𝑥

4𝑦' =− 16𝐶2𝑒

−4𝑥

𝑦'' = 16𝐶2𝑒

_______________
4𝑦' + 𝑦'' = 0
Thus,
𝑦'' + 4𝑦' = 0 Ans

ℎ
𝑚

10
...


Solution:
𝑦' = 𝑚 + 0
𝑚 = 𝑦'
Thus,
𝑦 = 𝑥𝑦' +

ℎ
𝑦'

Ans



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