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Magazine Chapter 6: Problem 61
Solve the first-order differential equation by any appropriate method. $$ \left(3 y^{2}+4 x y\right) d x+\left(2 x y+x^{2}\right) d y=0 $$
Short Answer
Expert verified
The solution of the given first-order differential equation is \( F(x, y) = 3y^2\log |x| + 4xy + y^2 = c \)
Step by step solution
01
Checking if the differential equation is exact
An exact differential equation has the form \( M(x, y) dx + N(x, y) dy = 0 \), and it is exact if \( \partial M/\partial y = \partial N/\partial x \). For this equation, we have \( M(x,y) = 3y^2+4xy \) and \( N(x,y) = 2xy+x^2 \). Therefore, \( \partial M/\partial y = 6y+4x \) and \( \partial N/\partial x = 2y+2x \). We can see that the two partial derivatives are not equal, so this specific equation is not exact.
02
Introduce integrating factor
We can turn an inexact differential equation into an exact one by introducing an integrating factor. In our case, a suitable integrating factor is \( m(x)=1/x \). We multiply entire equation by \( m(x) \): \( (3y^2/x + 4y) dx + (2y + x) dy = 0 \)
03
Check if it's now an exact differential equation
Let's recalculate the partial derivatives. The new \( M(x,y) = 3y^2/x + 4y \) and \( N(x,y) = 2y + x \). Therefore, \( \partial M/\partial y = 6y/x + 4 \) and \( \partial N/\partial x = 1 \). Now the two expressions are equal, so the equation is exact.
04
Solve the exact differential equation
The solution of an exact differential equation is the function \( F(x, y) = c \) which solves \( F_x = M \) and \( F_y = N \). We can integrate \( M \) with respect to \( x \) and \( N \) with respect to \( y \) to get two solutions which should agree: \( F_1 = \int (3y^2/x + 4y) dx = 3y^2\log |x| + 4xy + g_1(y) \) and \( F_2 = \int (2y + x) dy = y^2 + xy + g_2(x) \). Comparing the two, we find that \( g_1(y) = y^2 \) and \( g_2(x) = 3\log |x| \). Thus, the solution to the original equation is: \( F(x, y) = 3y^2\log |x| + 4xy + y^2 = c \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exact Differential Equation
An exact differential equation is one in which a given differential expression takes the form \( M(x, y)\, dx + N(x, y)\, dy = 0 \). A crucial property of an exact differential equation is that it satisfies \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
This implies that the function derivatives with respect to the other variable are equal, making certain algebraic methods applicable to solve for a potential function \( F(x, y) \).
This implies that the function derivatives with respect to the other variable are equal, making certain algebraic methods applicable to solve for a potential function \( F(x, y) \).
- When the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) holds true, the equation is said to be exact, and we can find a function, \( F(x, y) \), where its partial derivatives return \( M \) and \( N \) respectively.
- Solving the exact differential equation involves integrating \( M(x, y) \) with respect to \( x \), and \( N(x, y) \) with respect to \( y \). These integrals should agree
- If the initial equation isn't exact, as in our example, we can introduce an integrating factor to make it exact.
Integrating Factor
An integrating factor is a function used to transform a non-exact differential equation into an exact one. This is particularly useful when dealing with first-order differential equations that aren't exact.
In the given problem, the equation was not exact initially, so an integrating factor was introduced:
In the given problem, the equation was not exact initially, so an integrating factor was introduced:
- To determine an integrating factor, look for a function \( \mu(x) \) such that when multiplied by the whole differential equation, the new equation becomes exact.
- For the example equation, the integrating factor was \( \mu(x) = 1/x \). Multiplying the differential equation by this factor aids in achieving the condition for exactness.
- The role of the integrating factor is to alter the structures of \( M \) and \( N \) so that \( \frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x} \) becomes true.
First-order Differential Equation
A first-order differential equation is characterized by derivatives that are of the first degree. It involves only the first derivatives of the dependent variable with respect to one or more independent variables.
They are expressed in the form \( dy/dx = f(x, y) \) or in a differential form like \( M(x, y)\, dx + N(x, y)\, dy = 0 \).
They are expressed in the form \( dy/dx = f(x, y) \) or in a differential form like \( M(x, y)\, dx + N(x, y)\, dy = 0 \).
- Such equations might require different techniques to solve depending on their properties, such as separability or linearity.
- In our example, the differential equation required checking for exactness and using a suitable method like introducing an integrating factor to solve it.
- Mastering the basics of first-order differential equations can simplify understanding more complex systems in higher orders.
Partial Derivatives
Partial derivatives are a significant concept when dealing with multivariable functions in calculus. They represent the rate of change of a function with respect to one of its variables, keeping the other variables constant.
In the context of differential equations, partial derivatives play a vital role in assessing exactness:
In the context of differential equations, partial derivatives play a vital role in assessing exactness:
- The symbols \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) denote partial derivatives of \( M(x, y) \) and\( N(x, y) \) with respect to \( y \) and \( x \), respectively.
- They are crucial for determining whether a given equation is exact through the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
- Understanding how to compute and interpret partial derivatives is essential for solving differential equations, as they often provide insightful information about the variables鈥� interdependence.
