First-Order Exact DE - Form Definition
Click here to expand...
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:
Click here to expand...
- (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
Click here to expand...
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
Click here to expand...
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
- first we integrate 𝑀(𝑥,𝑦) with respect to 𝑑𝑥
- ∫𝑀(𝑥,𝑦)𝑑𝑥 = 𝐹(𝑥,𝑦) + 𝑔(𝑦)
- we know that 𝐹(𝑥,𝑦) + 𝑔(𝑦)equals the unknown function 𝜑(𝑥,𝑦)
- 𝜑(𝑥,𝑦) = 𝐹(𝑥,𝑦) + 𝑔(𝑦)
- but we don’t know what 𝑔(𝑦) is, so we need to solve for it:
- we know that:
- 𝑑𝜑/𝑑𝑦 = 𝑁(𝑥,𝑦)
- 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)
- once the equation (𝛿/𝛿𝑦)(𝐹(𝑥,𝑦) + 𝑔(𝑦)) = 𝑁(𝑥,𝑦) is computed we solve for 𝑔(𝑦)
- next we come back to the equation 𝐹(𝑥,𝑦) + 𝑔(𝑦) and plugin what we found at step 3.c for 𝑔(𝑦):
- 𝜑(𝑥,𝑦) = 𝐹(𝑥,𝑦) + 𝑔(𝑦)
- 𝜑(𝑥,𝑦) = 𝐹(𝑥,𝑦) + (thing from 3.c)
- next we take the original DE
- now we merge (4.b) and (5.f) and get:
- 𝐹(𝑥,𝑦) + (thing from 3.c) = 𝐶
- next we apply the initial conditions to solve for 𝐶
- once we solve for 𝐶 we plug the scalar back into:
- 𝐹(𝑥,𝑦) + (thing from 3.c) = 𝑠𝑐𝑎𝑙𝑎𝑟-𝐶
- 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
- first we integrate 𝑀(𝑥,𝑦) with respect to 𝑑𝑥
- ∫𝑀(𝑥,𝑦)𝑑𝑥 = ∫(2𝑥+𝑦2)𝑑𝑥 = 𝑥2 + 𝑥𝑦2 + 𝑔(𝑦)
- we know that 𝑥2 + 𝑥𝑦2 + 𝑔(𝑦) equals the unknown function 𝜑(𝑥,𝑦)
- 𝜑(𝑥,𝑦) = 𝑥2 + 𝑥𝑦2 + 𝑔(𝑦)
- but we don’t know what 𝑔(𝑦) is, so we need to solve for it:
- we know that:
- 𝑑𝜑/𝑑𝑦 = 𝑁(𝑥,𝑦) = 2𝑥𝑦
- 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
- thus we get:
- 𝑔’(𝑦) = 0
- 𝑑𝑔(𝑦)/𝑑𝑦 = 0
- ∫𝑑𝑔(𝑦) = ∫0·𝑑𝑦
- 𝑔(𝑦) = 0 + 𝑐 = 𝑐
- 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
- now we know that:
- 𝜑(𝑥,𝑦) = 𝑥2 + 𝑥𝑦2= 𝐶
- next we apply the initial condition 𝑦(1) = 2 to solve for 𝐶
- 12 + 1·22= 𝐶
- 𝐶 = 5
- the solution is then:
- 𝑥2 + 𝑥𝑦2= 𝐶
- 𝑥2 + 𝑥𝑦2= 5
- 𝑥2 + 𝑥𝑦2- 5 = 0
- 𝑥𝑦2+ 𝑥2- 5 = 0
- solve for 𝑦:
- 𝑦 = [(5 - 𝑥2)/𝑥](1/2)
- or using the quadratic equation:
- 𝑥2 + 𝑥𝑦2= 5
- 𝑥2 + 𝑥𝑦2- 5 = 0
- 𝑥𝑦2+ 𝑥2- 5 = 0
- where:
- 𝑎 = 𝑥
- 𝑏 = 0
- 𝑐 = (𝑥2- 5)
- then:
- 𝑦 = [±√(-4𝑥(𝑥2-5))] / [2𝑥]
- 𝑦 = [±√(-4𝑥3+ 20𝑥)] / [2𝑥]
- hence 𝑦 equals either:
- 𝑦 = √(-4𝑥3+ 20𝑥) / (2𝑥)
- 𝑦 = -√(-4𝑥3+ 20𝑥) / (2𝑥)