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Exact Differential Equation
An exact differential equation is a type of first-order ordinary differential equation (ODE) that can be written in a specific form and has a special property. In an exact differential equation, the total differential of an unknown function can be expressed as the exact derivative of some other function. This property allows for a straightforward method of solving such equations.
The general form of an exact differential equation is as follows:
M(x, y)dx + N(x, y)dy = 0
Here, M(x, y) and N(x, y) are functions of two variables, x and y. To check if an ODE is exact, you can use the following criterion:
∂M/∂y = ∂N/∂x
If this condition is satisfied, the equation is exact. If not, it’s not an exact differential equation.
To solve an exact differential equation, you can follow these steps:
Check if the equation is exact using the criterion above.
If it’s exact, find a function F(x, y) such that:
∂F/∂x = M(x, y)
∂F/∂y = N(x, y)
Once you’ve found F(x, y), the solution to the equation can be obtained by writing F(x, y) = C, where C is the constant of integration.
The solution C implicitly defines the relationship between x and y, and it may not be in a more explicit form unless additional initial or boundary conditions are given.
Solving exact differential equations can be a powerful technique in various areas of mathematics and science, including physics and engineering, where many real-world problems are described by these types of equations.
Proof of the Theorem
Here is the proof of the theorem related to exact differential equations using normal alphabetical characters:
Definition: A first-order differential equation of the form
- M(x, y)dx + N(x, y)dy = 0
is said to be an exact differential equation if there exists a continuously differentiable function ψ(x, y) such that
∂ψ/∂x = M(x, y)
∂ψ/∂y = N(x, y)
In other words, the equation is exact if the partial derivatives of some function ψ(x, y) with respect to x and y are equal to M(x, y) and N(x, y), respectively.
Theorem: If a differential equation M(x, y)dx + N(x, y)dy = 0 is exact, then it has a solution of the form ψ(x, y) = C, where C is a constant.
Proof:
Assume that the given differential equation is exact, meaning there exists a function ψ(x, y) such that ∂ψ/∂x = M(x, y) and ∂ψ/∂y = N(x, y).
Consider the first partial derivative of ψ(x, y) with respect to x:
∂ψ/∂x = M(x, y)
Now, integrate both sides of this equation with respect to x:
∫ ∂ψ/∂x dx = ∫ M(x, y) dx
Using the fundamental theorem of calculus, we get:
ψ(x, y) = ∫ M(x, y) dx + g(y)
Where g(y) is an arbitrary function of y.
Next, consider the partial derivative of ψ(x, y) with respect to y:
∂ψ/∂y = N(x, y)
Now, differentiate the expression for ψ(x, y) obtained in step 2 with respect to y:
d/dy ( ∫ M(x, y) dx + g(y) ) = N(x, y)
Since ∫ M(x, y) dx does not depend on y, its derivative with respect to y is zero, and we are left with:
g'(y) = N(x, y)
To find g(y), integrate both sides with respect to y:
∫ g'(y) dy = ∫ N(x, y) dy
This gives us:
g(y) = ∫ N(x, y) dy + C₁
Where C₁ is an arbitrary constant.
Combining the expressions for ψ(x, y) obtained in step 2 and step 4, we get:
ψ(x, y) = ∫ M(x, y) dx + ∫ N(x, y) dy + C₁
Define a new constant C such that C = C₁ + ψ(x₀, y₀), where (x₀, y₀) is a particular point in the solution.
Finally, we have:
ψ(x, y) = ∫ M(x, y) dx + ∫ N(x, y) dy + C
This is the general solution to the exact differential equation, and it is of the form ψ(x, y) = C, where C is a constant. This completes the proof of the theorem.
Exact Differential Equation Formula
An exact differential equation is a type of first-order differential equation that can be written in the form:
M(x, y)dx + N(x, y)dy = 0
Where:
M(x, y) and N(x, y) are functions of two variables, x and y.
dx and dy represent the differentials of x and y, respectively.
The exact differential equation has a special property: it can be derived from a function called a potential function, also known as a “potential” or “integrating factor” (though this term can have other meanings in different contexts). This potential function is usually denoted as U(x, y).
The relationship between the potential function U and the functions M and N in the exact differential equation is given by the following two partial derivatives:
∂U/∂x = M
∂U/∂y = N
If these partial derivatives exist and satisfy these conditions, the given differential equation is exact.
Solving an exact differential equation typically involves finding the potential function U(x, y) and then using it to obtain the solution. Here are the general steps to solve an exact differential equation:
Check if the equation is exact by verifying if ∂U/∂x = M and ∂U/∂y = N.
If the equation is exact, integrate ∂U/∂x with respect to x to find U(x, y) up to an arbitrary constant C1:
∫M dx = U(x, y) + C1
Integrate ∂U/∂y with respect to y using the potential function U(x, y) obtained in step 2, up to another arbitrary constant C2:
∫N dy = U(x, y) + C2
Set the expressions obtained in steps 2 and 3 equal to each other:
U(x, y) + C1 = U(x, y) + C2
Solve for C2 in terms of C1:
C2 = C1 + constant (let’s call it C)
Express U(x, y) as a function of x, y, and C:
U(x, y) = C – C1
The general solution to the exact differential equation is:
U(x, y) = C – C1
Where C and C1 are arbitrary constants.
Keep in mind that solving specific exact differential equations may involve additional techniques and steps, but this is the general approach for solving them.
How to Solve Exact Differential Equation?
To solve an exact differential equation, you need to follow a systematic approach that involves identifying whether the equation is exact, finding an integrating factor if necessary, and then integrating to find the solution. Here are the steps:
-
Identify the Equation as Exact: An exact differential equation has the form: M(x, y)dx + N(x, y)dy = 0 Where M and N are functions of two variables, x and y.
The equation is exact if the following condition is met: ∂M/∂y = ∂N/∂x
If this condition is satisfied, then you have an exact differential equation, and you can proceed to the next steps. If not, you’ll need to find an integrating factor to make it exact.
-
Find the Integrating Factor (if needed): If the equation is not initially exact (∂M/∂y ≠ ∂N/∂x), you can make it exact by multiplying both sides of the equation by an integrating factor, μ(x, y), which is a function of x and y. The integrating factor μ(x, y) is chosen to make the equation exact, and it should satisfy the following condition: μ(x, y)[∂M/∂y – ∂N/∂x] = ∂μ/∂y
Rearrange this equation to solve for μ(x, y): μ(x, y) = e^∫[∂μ/∂y]dy
Once you find μ(x, y), multiply both sides of the original equation by μ(x, y): μ(x, y)[M(x, y)dx + N(x, y)dy] = 0
Now, the equation is exact.
-
Integrate to Find the Solution: With the equation in exact form, you can find the solution by integrating both sides with respect to x and y separately:
∫[μ(x, y)M(x, y)]dx + ∫[μ(x, y)N(x, y)]dy = C
Where C is the constant of integration.
-
Simplify and Solve for y: If necessary, you may need to simplify the equation further and solve for y. This may involve rearranging terms and performing the integrations.
-
Solve for the Constant of Integration: If you have boundary conditions or initial conditions, you can use them to solve for the constant of integration, C.
-
Write the Final Solution: Write the solution in the form y = f(x) if possible or as an implicit equation in terms of x and y.
It’s important to note that exact differential equations can sometimes be quite complex, and the choice of the integrating factor can require some creativity and skill. Additionally, the process of finding the integrating factor and integrating the equation can be laborious for certain equations. Practice and familiarity with different techniques are essential for solving these types of equations effectively.
Exact Differential Equation Examples
An exact differential equation is a type of first-order ordinary differential equation that can be written in the form:
M(x, y) dx + N(x, y) dy = 0
Where M and N are functions of two variables (x and y) and have continuous partial derivatives. An exact differential equation has a unique solution when it can be expressed as the total differential of some function u(x, y):
du = M dx + N dy
When du can be expressed this way, the equation is said to be exact. Here are a couple of examples of exact differential equations:
Example 1:
(2x + 3y) dx + (x^2 + 2y) dy = 0
To check if it’s exact, we calculate the partial derivatives of M and N with respect to y and x, respectively:
∂M/∂y = 3
∂N/∂x = 2x
Since ∂M/∂y is not equal to ∂N/∂x, it’s not exact.
Example 2:
(2xy + y^2) dx + (x^2 + 2xy) dy = 0
To check if it’s exact, we calculate the partial derivatives of M and N with respect to y and x, respectively:
∂M/∂y = 2x + 2y
∂N/∂x = 2x + 2y
In this case, ∂M/∂y = ∂N/∂x, so it’s exact. To find the solution, we need to find a function u(x, y) such that du = (2xy + y^2) dx + (x^2 + 2xy) dy. We can do this by integrating each term:
∫(2xy + y^2) dx = x^2y + (1/3)y^3 + h(y)
Now, we find the partial derivative of this expression with respect to y and set it equal to N:
∂/∂y [x^2y + (1/3)y^3 + h(y)] = x^2 + y^2 + h'(y) = x^2 + 2xy
Solving for h'(y):
h'(y) = y^2
Integrating h'(y) with respect to y:
h(y) = (1/3)y^3 + C
So, the general solution is:
x^2y + (1/3)y^3 + (1/3)y^3 + C = C
These are examples of exact differential equations and how to solve them. In the first example, the equation was not exact, while in the second example, it was exact, and we found the general solution.
Solved Examples on Exponential Function
Here are some solved examples involving exponential functions:
Example 1:
Find the value of y if y = 2^3.
Solution:
y = 2^3
y = 2 * 2 * 2
y = 8
So, the value of y is 8.
Example 2:
Solve for x in the equation e^x = 20.
Solution:
Take the natural logarithm (ln) of both sides to solve for x:
ln(e^x) = ln(20)
Using the property that ln(e^x) = x:
x = ln(20)
Using a calculator, approximate the value of x:
x ≈ 2.9957
So, x is approximately equal to 2.9957.
Example 3:
A population of bacteria doubles every hour. If there are initially 100 bacteria, how many bacteria will there be after 5 hours?
Solution:
We can model the population growth with an exponential function:
P
Where:
P
P₀ is the initial population (100 in this case).
t is the time in hours.
Plugging in the values:
P(5) = 100 * 2^5
P(5) = 100 * 32
P(5) = 3200
So, there will be 3200 bacteria after 5 hours.
Example 4:
A car depreciates at a rate of 15% per year. If the car is initially worth $20,000, what will its value be after 3 years?
Solution:
We can model the depreciation with an exponential function:
V
Where:
V
V₀ is the initial value ($20,000 in this case).
r is the annual depreciation rate (15% or 0.15 as a decimal).
t is the number of years (3 years in this case).
Plugging in the values:
V(3) = 20000 * (1 – 0.15)^3
V(3) = 20000 * (0.85)^3
V(3) ≈ 20000 * 0.6141
V(3) ≈ $12,282.00
So, the value of the car after 3 years will be approximately $12,282.00.
These examples demonstrate various applications of exponential functions in different contexts.
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