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Magazine Chapter 1: Problem 33
Verify that the given differential equation is exact; then solve it. $$ \left(3 x^{2}+2 y^{2}\right) d x+\left(4 x y+6 y^{2}\right) d y=0 $$
Short Answer
Expert verified
The differential equation is exact and its solution is \( x^3 + 2y^2x + 2y^3 = C \).
Step by step solution
01
Define the functions for exactness verification
In the given differential equation, identify the functions \( M(x, y) = 3x^2 + 2y^2 \) and \( N(x, y) = 4xy + 6y^2 \). The equation is in the form \( M(x, y) \, dx + N(x, y) \, dy = 0 \).
02
Calculate partial derivatives for exactness
Calculate the partial derivative of \( M \) with respect to \( y \), \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2 + 2y^2) = 4y \). Similarly, calculate the partial derivative of \( N \) with respect to \( x \), \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(4xy + 6y^2) = 4y \).
03
Verify exactness condition
Check if the partial derivatives from Step 2 are equal: \( \frac{\partial M}{\partial y} = 4y \) and \( \frac{\partial N}{\partial x} = 4y \). Since they are equal, the differential equation is exact.
04
Integrate M(x, y) to find F(x, y)
Integrate the function \( M(x, y) = 3x^2 + 2y^2 \) with respect to \( x \): \[ \int (3x^2 + 2y^2) \, dx = x^3 + 2y^2x + g(y) \] where \( g(y) \) is an arbitrary function of \( y \).
05
Differentiate F(x, y) with respect to y and equate to N(x, y)
Differentiate \( F(x, y) = x^3 + 2y^2x + g(y) \) with respect to \( y \): \[ \frac{\partial F}{\partial y} = 4xy + g'(y) \] Set this equal to \( N(x, y) = 4xy + 6y^2 \): \[ 4xy + g'(y) = 4xy + 6y^2 \] Thus, \( g'(y) = 6y^2 \).
06
Solve for g(y) by integrating g'(y)
Integrate \( g'(y) = 6y^2 \) with respect to \( y \) to find \( g(y) \): \[ g(y) = \int 6y^2 \, dy = 2y^3 + C \] where \( C \) is a constant of integration.
07
Write the solution F(x, y) = constant
Substitute \( g(y) = 2y^3 + C \) into \( F(x, y) \): \[ F(x, y) = x^3 + 2y^2x + 2y^3 = C \] Therefore, the general solution to the differential equation is: \[ x^3 + 2y^2x + 2y^3 = C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept when dealing with functions of multiple variables. In mathematics, the partial derivative of a function with respect to one variable is the derivative considering all other variables constant.
This is particularly useful in exact differential equations, where you want to ensure a condition known as exactness.
This is particularly useful in exact differential equations, where you want to ensure a condition known as exactness.
- To find the partial derivative of a function with respect to a variable, differentiate it by treating all other variables as constants.
- For example, if you have a function like \( M(x, y) = 3x^2 + 2y^2 \), the partial derivative of \( M \) with respect to \( y \) is the derivative of only the terms involving \( y \), resulting in \( 4y \).
- Similarly, for \( N(x, y) = 4xy + 6y^2 \), the partial derivative with respect to \( x \) simplifies to \( 4y \).
Integration
Integration is the process of finding the integral, which is effectively the opposite of taking a derivative. In solving exact differential equations, integration helps to construct a potential function \( F(x, y) \) that remains constant over equal waves (contour lines) of the equation.
- Given the function \( M(x, y) = 3x^2 + 2y^2 \), integrating with respect to \( x \) while considering \( y \) as a constant yields \( x^3 + 2y^2x \).
- The integration adds an unknown function \( g(y) \) to account for any function of \( y \) alone, which doesn't affect the derivative with respect to \( x \).
- Once you have this integral, the next step is to differentiate it with respect to \( y \) and match it with \( N(x, y) \) to solve for \( g(y) \).
Exactness Condition
The exactness condition is a criterion used to verify that a differential equation is exact and therefore solvable through the integration process.
The condition checks the equality of mixed partial derivatives.
The condition checks the equality of mixed partial derivatives.
- In an exact differential equation of form \( M(x, y)\, dx + N(x, y)\, dy = 0 \), exactness requires that \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
- For our equation, we verified this by showing the partial derivatives \( \frac{\partial M}{\partial y} = 4y \) and \( \frac{\partial N}{\partial x} = 4y \) are equal.
- When this condition is met, it confirms there is a scalar potential function \( F(x, y) \) such that \( dF = M dx + N dy \), leading to the resolve of the equation.
General Solution
The general solution involves finding a function that satisfies the given exact differential equation, encompassing all possibilities over a constant.
- Once the potential function \( F(x, y) \) is determined, setting \( F(x, y) = C \) ((where \( C \) is a constant) provides the structure of the solution.
- For instance, the equation \( x^3 + 2y^2x + 2y^3 = C \) describes all the curves on which the differential equation holds true.
- This general solution is achieved by formulating the entire function including adjustments for additional functions like \( g(y) \), ensuring all parts match the original equation.
