Problem 31 - Advanced Engineering Mathematics

Question

Solve the given differential equation by finding, as in Example 4 , an appropriate integrating factor. $$ \left(2 y^{2}+3 x\right) d x+2 x y d y=0 $$

Step-by-Step Solution

Answer

The solution is \( x^2 e^{y^2} + 2xy^2 e^{y^2} = C \).

Step 1: Identify the Form

First, we need to identify whether the given differential equation is in the form of a linear differential equation. The equation is \( (2y^2 + 3x) dx + 2xy dy = 0 \). This is a first-order differential equation that is not exact. We have \( M(x, y) = 2y^2 + 3x \) and \( N(x, y) = 2xy \).

Step 2: Check for Exactness

To check for exactness, we compare the partial derivatives: \( \frac{\partial M}{\partial y} = \frac{\partial (2y^2 + 3x)}{\partial y} = 4y \) and \( \frac{\partial N}{\partial x} = \frac{\partial (2xy)}{\partial x} = 2y \). Since \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \), the equation is not exact.

Step 3: Determine the Integrating Factor

Since the differential equation is not exact, we will look for an integrating factor. We use the formula for cases where \( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = f(x) \): the integrating factor is \( \mu(y) = e^{\int f(x) \, dy } \). In this case, \( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = 4y - 2y = 2y \), so \( \mu(y) = e^{\int 2y \, dy } = e^{y^2} \).

Step 4: Multiply and Simplify

Now, multiply the entire differential equation by the integrating factor \( e^{y^2} \): \[ e^{y^2}((2y^2 + 3x) dx + 2xy dy) = 0 \]. The equation becomes \( ( e^{y^2}(2y^2 + 3x) ) dx + ( e^{y^2}(2xy) ) dy = 0 \).

Step 5: Check and Solve as an Exact Equation

Check the new equation for exactness with \( \tilde{M} = e^{y^2}(2y^2 + 3x) \) and \( \tilde{N} = e^{y^2}(2xy) \). Find partial derivatives: \( \frac{\partial \tilde{M}}{\partial y} = e^{y^2}(4y^3 + 4xy + 6y) \) and \( \frac{\partial \tilde{N}}{\partial x} = e^{y^2}(2y) \). These are equal, so the equation is exact. Integrate \( \tilde{M} \) with respect to \( x \) to find \( \phi(x,y) = e^{y^2}(x^2 + x(2y^2)) + h(y) \).

Step 6: Solve for the Function \( \phi(x, y) \)

Integrate \( \tilde{N} \) with respect to \( y \) and equate it with the derivative of \( \phi(x, y) \) to find \( h(y) \). This gives us the solution: \[ x^2 e^{y^2} + 2xy^2 e^{y^2} = C \], as \( h(y) \) is constant because all terms incorporate \( y \).

Step 7: Conclusion

Thus, the solution to the differential equation is \( x^2 e^{y^2} + 2xy^2 e^{y^2} = C \), where \( C \) is a constant determined by initial conditions if provided.

Key Concepts

Integrating Factor

When solving differential equations, especially non-exact ones, an integrating factor can be a powerful tool. Imagine an integrating factor as a magic multiplier that transforms a complicated equation into something more straightforward. In our exercise, the differential equation wasn't exact, which means the standard methods of solving it don't apply directly. Here's where the integrating factor comes into play.

• An integrating factor is defined for a differential equation to convert it into an exact equation, making it solvable through integration.
• It is a function, often denoted by \( \mu \), that when multiplied with the original equation, balances the scales between two differential components.
• In the equation \((2y^2 + 3x) dx + 2xy dy = 0\), the factor is chosen based on the comparison of partial derivatives.

In our particular solution, we found that the integrating factor was \( e^{y^2} \), achieved by integrating the difference of the derivatives with respect to \( y \). This balancing trick allows us to rewrite our equation, making integration much more manageable.

Exact Differential Equation

An exact differential equation is a special type of equation where there's a "perfect match" between its parts. In simple terms, if you have two functions involved, like \( M(x, y) \) and \( N(x, y) \), they fit together just right.

• In mathematical language, an equation is exact if the cross derivatives are equal: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
• This characteristic means both parts can be combined into a single integrable function, \( \phi(x, y) \).
• If not exact initially, like in our example, making it exact with an integrating factor allows for integration.

Why does being exact matter? Because once it's exact, you can find a function \( \phi \) such that its total derivative matches the original differential equation, leading you to the solution directly and efficiently.

First-Order Differential Equation

A first-order differential equation is one of the simplest types of differential equations. It involves derivatives of the first degree, meaning only the first derivative of the function and perhaps the function itself appear in the equation. These equations are common across various scientific fields, modeling exponential growth, cooling laws, and circuits.

• They're typically represented as \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are known functions of \( x \).
• Our example equation \((2y^2 + 3x) dx + 2xy dy = 0\) also is first-order, involving no higher derivatives.
• Methods for solving include separation of variables, linear equation solving, and using integrating factors, like we did here.

First-order equations are foundational; understanding them is crucial because they form the basis for understanding more complex equations. Often, being able to recognize one is half the battle in differential equation solving.

Partial Derivatives

In multivariable calculus, partial derivatives are a way to explore changes of functions with several variables. If you picture a mountain as a surface, taking a partial derivative is like finding the slope of this mountain in one specific direction.

• Partial derivatives, denoted as \( \frac{\partial}{\partial x} \) or \( \frac{\partial}{\partial y} \), measure how a function changes as only one variable is varied, keeping others constant.
• In our exercise, partial derivatives were crucial for checking exactness. We compared \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) to decide if our equation was exact.
• These derivatives helped in determining the need for an integrating factor and enabled the transformation of our differential equation into a solvable form.

By understanding partial derivatives, we delve into how functions behave in a multidimensional space, which is vital for solving differential equations involving several variables.