First-Order Exact DE

First-Order Exact DE - Form Definition

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a first-order differential equation

• 𝑀(𝑥,𝑦) + 𝑁(𝑥,𝑦)(𝑑𝑦/𝑑𝑥) = 0

is of exact form if there exist an unknown function 𝜑(𝑥,𝑦) such that:

• (𝛿𝜑/𝛿𝑥) = 𝑀(𝑥,𝑦)
• (𝛿𝜑/𝛿𝑦) = 𝑁(𝑥,𝑦)

for more details:

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• (𝛿/𝛿𝑥) [𝜑(𝑥,𝑦)] = 0
• (𝛿/𝛿𝑥)𝜑(𝑥,𝑦) + (𝛿/𝛿𝑦)𝜑(𝑥,𝑦)·(𝑑𝑦/𝑑𝑥) = 0
• (𝛿/𝛿𝑥)𝜑(𝑥,𝑦) + (𝛿/𝛿𝑦)𝜑(𝑥,𝑦)·(𝑑𝑦/𝑑𝑥) = 0
• (𝛿𝜑/𝛿𝑥) + (𝛿𝜑/𝛿𝑦)·(𝑑𝑦/𝑑𝑥) = 0

now we want:

• (𝛿𝜑/𝛿𝑥) = 𝑀(𝑥,𝑦)
• (𝛿𝜑/𝛿𝑦) = 𝑁(𝑥,𝑦)

First-Order Exact DE - Form Verification

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a first-order differential equation

• 𝑀(𝑥,𝑦) + 𝑁(𝑥,𝑦)(𝑑𝑦/𝑑𝑥) = 0

assume that there is an unknown function 𝜑(𝑥,𝑦) such that:

• 𝑀(𝑥,𝑦) = (𝛿𝜑/𝛿𝑥)
• 𝑁(𝑥,𝑦) = (𝛿𝜑/𝛿𝑦)

from Calculus we know that the mixed partials are equal, in other words:

• (𝛿/𝛿𝑦)(𝛿𝜑/𝛿𝑥) = (𝛿/𝛿𝑥)(𝛿𝜑/𝛿𝑦)

thus we can substitute and get:

• (𝛿/𝛿𝑦)𝑀(𝑥,𝑦) = (𝛿/𝛿𝑥)𝑁(𝑥,𝑦)

now we plugin the functions for 𝑀(𝑥,𝑦) and 𝑁(𝑥,𝑦) then calculate the partial derivatives on each side. If they are equal then the DE is of exact form

First-Order Exact DE - Solving It Method

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based on: TheLazyEngineer - YouTube

given a first-order differential equation:

• 𝑀(𝑥,𝑦) + 𝑁(𝑥,𝑦)·𝑦’ = 0

and initial condition:

• 𝑦(𝑥) = ?

we first verify that it is of exact form (detailed steps shown above)

• we need to verify that the following holds:
  • (𝛿/𝛿𝑦)𝑀(𝑥,𝑦) = (𝛿/𝛿𝑥)𝑁(𝑥,𝑦)
• substitute 𝑀(𝑥,𝑦) & 𝑁(𝑥,𝑦), caculate partial derivatives, then verify above equation holds
• if it hold then the DE is of exact form

if the DE is of exact form we could start to solve it

1. first we integrate 𝑀(𝑥,𝑦) with respect to 𝑑𝑥

• ∫𝑀(𝑥,𝑦)𝑑𝑥 = 𝐹(𝑥,𝑦) + 𝑔(𝑦)

2. we know that 𝐹(𝑥,𝑦) + 𝑔(𝑦)equals the unknown function 𝜑(𝑥,𝑦)

• 𝜑(𝑥,𝑦) = 𝐹(𝑥,𝑦) + 𝑔(𝑦)

3. but we don’t know what 𝑔(𝑦) is, so we need to solve for it:
4. we know that:
   • 𝑑𝜑/𝑑𝑦 = 𝑁(𝑥,𝑦)
5. thus we differentiate 𝐹(𝑥,𝑦) + 𝑔(𝑦) with respect to 𝑦 and set it equal to 𝑁(𝑥,𝑦)
   • 𝑑𝜑/𝑑𝑦 = 𝑁(𝑥,𝑦)
   • (𝛿/𝛿𝑦)(𝐹(𝑥,𝑦) + 𝑔(𝑦)) = 𝑁(𝑥,𝑦)
6. once the equation (𝛿/𝛿𝑦)(𝐹(𝑥,𝑦) + 𝑔(𝑦)) = 𝑁(𝑥,𝑦) is computed we solve for 𝑔(𝑦)
7. next we come back to the equation 𝐹(𝑥,𝑦) + 𝑔(𝑦) and plugin what we found at step 3.c for 𝑔(𝑦):
8. 𝜑(𝑥,𝑦) = 𝐹(𝑥,𝑦) + 𝑔(𝑦)
9. 𝜑(𝑥,𝑦) = 𝐹(𝑥,𝑦) + (thing from 3.c)
10. next we take the original DE
11. now we merge (4.b) and (5.f) and get:
12. 𝐹(𝑥,𝑦) + (thing from 3.c) = 𝐶
13. next we apply the initial conditions to solve for 𝐶
14. once we solve for 𝐶 we plug the scalar back into:
15. 𝐹(𝑥,𝑦) + (thing from 3.c) = 𝑠𝑐𝑎𝑙𝑎𝑟-𝐶
16. then we solve for 𝑦

First-Order Exact DE - Solving It (Examples)

Example 1

based on: TheLazyEngineer - YouTube

given a first-order differential equation:

• 2𝑥+𝑦² + 2𝑥𝑦·𝑦’ = 0 # 𝑀(𝑥,𝑦) + 𝑁(𝑥,𝑦)·𝑦’ = 0

and initial condition:

• 𝑦(1) = 2

we first verify that it is of exact form

• we need to verify that the following holds:
  • (𝛿/𝛿𝑦)𝑀(𝑥,𝑦) = (𝛿/𝛿𝑥)𝑁(𝑥,𝑦)
• substitute and verify:
  • (𝛿/𝛿𝑦)𝑀(𝑥,𝑦) = (𝛿/𝛿𝑥)𝑁(𝑥,𝑦)
  • (𝛿/𝛿𝑦)2𝑥+𝑦² = (𝛿/𝛿𝑥)2𝑥𝑦
  • 2𝑦 = 2𝑦
• it holds, thus 2𝑥+𝑦² + 2𝑥𝑦·𝑦’ = 0 is of exact form

if the DE is of exact form we could start to solve it

1. first we integrate 𝑀(𝑥,𝑦) with respect to 𝑑𝑥

• ∫𝑀(𝑥,𝑦)𝑑𝑥 = ∫(2𝑥+𝑦²)𝑑𝑥 = 𝑥² + 𝑥𝑦² + 𝑔(𝑦)

2. we know that 𝑥² + 𝑥𝑦² + 𝑔(𝑦) equals the unknown function 𝜑(𝑥,𝑦)

• 𝜑(𝑥,𝑦) = 𝑥² + 𝑥𝑦² + 𝑔(𝑦)

3. but we don’t know what 𝑔(𝑦) is, so we need to solve for it:
4. we know that:
   • 𝑑𝜑/𝑑𝑦 = 𝑁(𝑥,𝑦) = 2𝑥𝑦
5. thus we differentiate 𝑥² + 𝑥𝑦² + 𝑔(𝑦) with respect to 𝑦 and set it equal to 2𝑥𝑦
   • 𝑑𝜑/𝑑𝑦 = 𝑁(𝑥,𝑦) = 2𝑥𝑦
   • (𝛿/𝛿𝑦)(𝑥² + 𝑥𝑦² + 𝑔(𝑦)) = 2𝑥𝑦
   • 2𝑥𝑦 + 𝑔’(𝑦) = 2𝑥𝑦
6. thus we get:
   • 𝑔’(𝑦) = 0
   • 𝑑𝑔(𝑦)/𝑑𝑦 = 0
   • ∫𝑑𝑔(𝑦) = ∫0·𝑑𝑦
   • 𝑔(𝑦) = 0 + 𝑐 = 𝑐
7. next we come back to the equation 𝑥² + 𝑥𝑦² + 𝑔(𝑦) and plugin 𝑐 for 𝑔(𝑦):

• 𝑥² + 𝑥𝑦² + 𝑔(𝑦) = 𝐶
• 𝑥² + 𝑥𝑦² + 𝑐 = 𝐶
• 𝑥² + 𝑥𝑦²= 𝐶

5. now we know that:
6. 𝜑(𝑥,𝑦) = 𝑥² + 𝑥𝑦²= 𝐶
7. next we apply the initial condition 𝑦(1) = 2 to solve for 𝐶
8. 1² + 1·2²= 𝐶
9. 𝐶 = 5
10. the solution is then:
11. 𝑥² + 𝑥𝑦²= 𝐶
12. 𝑥² + 𝑥𝑦²= 5
13. 𝑥² + 𝑥𝑦²- 5 = 0
14. 𝑥𝑦²+ 𝑥²- 5 = 0
15. solve for 𝑦:
16. 𝑦 = [(5 - 𝑥²)/𝑥]^(1/2)
17. or using the quadratic equation:
18. 1. 𝑥² + 𝑥𝑦²= 5
    2. 𝑥² + 𝑥𝑦²- 5 = 0
    3. 𝑥𝑦²+ 𝑥²- 5 = 0
19. where:
    1. 𝑎 = 𝑥
    2. 𝑏 = 0
    3. 𝑐 = (𝑥²- 5)
20. then:
    1. 𝑦 = [±√(-4𝑥(𝑥²-5))] / [2𝑥]
    2. 𝑦 = [±√(-4𝑥³+ 20𝑥)] / [2𝑥]
21. hence 𝑦 equals either:
    1. 𝑦 = √(-4𝑥³+ 20𝑥) / (2𝑥)
    2. 𝑦 = -√(-4𝑥³+ 20𝑥) / (2𝑥)