Problem 15 - Advanced Engineering Mathematics

Question

Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(x^{2} y^{3}-\frac{1}{1+9 x^{2}}\right) \frac{d x}{d y}+x^{3} y^{2}=0 $$

Step-by-Step Solution

Answer

The differential equation is not exact, so it cannot be solved by this method.

Step 1: Identify M and N

The given differential equation is \( \left( x^2 y^3 - \frac{1}{1+9x^2} \right) \frac{dx}{dy} + x^3 y^2 = 0 \). This can be rewritten in the form \( M(x, y) + N(x, y) \frac{dy}{dx} = 0 \), which means \( M(x, y) = x^3 y^2 \) and \( N(x, y) = x^2 y^3 - \frac{1}{1+9x^2} \).

Step 2: Compute \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \)

Compute the partial derivatives: \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (x^3 y^2) = 2x^3 y \) and \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left(x^2 y^3 - \frac{1}{1+9x^2}\right) = 2x y^3 + \frac{18x}{(1+9x^2)^2} \).

Step 3: Check for Exactness

The equation is exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Comparing both derivatives, \( 2x^3 y eq 2x y^3 + \frac{18x}{(1+9x^2)^2} \), showing the differential equation is not exact.

Key Concepts

Partial Derivatives

In calculus, a partial derivative represents the rate at which a function changes as one of its variables is altered, while keeping the other variables constant. This is an essential concept when dealing with functions of multiple variables, such as those in exact differential equations. In our specific exercise, we need the partial derivatives of functions \( M(x, y) \) and \( N(x, y) \).- For \( M(x, y) = x^3 y^2 \), the partial derivative with respect to \( y \) is computed as \( \frac{\partial M}{\partial y} = 2x^3 y \).- For \( N(x, y) = x^2 y^3 - \frac{1}{1+9x^2} \), the partial derivative with respect to \( x \) is \( \frac{\partial N}{\partial x} = 2x y^3 + \frac{18x}{(1+9x^2)^2} \).By understanding how to find these derivatives, we can further explore the behavior of the function in specific directions, helping us to determine the exactness of the differential equation.

Exactness Condition

The exactness condition is a critical test to determine if a given differential equation is exact. If a differential equation meets this condition, it implies that there exists a function \( \psi(x, y) \) such that the total derivative \( d\psi = M(x, y) + N(x, y) \frac{dy}{dx} = 0 \) reflects a potential function.To check the exactness of a differential equation, we need to compare the partial derivatives. Specifically, the condition \[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]must hold true for the equation to be exact.In our exercise, this was examined by comparing \( 2x^3 y \) to \( 2xy^3 + \frac{18x}{(1+9x^2)^2} \). Since they aren't equal, the differential equation is not exact, and it cannot be solved through this exactness method.

Differential Equation Solution

When dealing with exact differential equations, the goal is often to find a solution in the form of some potential function \( \psi(x, y) \) whose total derivative equates to zero. Solving these requires determining the precise relationship between \( x \) and \( y \) that satisfies the equation. Since our particular problem was not exact, direct solutions involving integral calculus and finding \( \psi(x, y) \) were not applicable here. However, if exact, the process typically involves:

• Integrating \( M(x, y) \) with respect to \( x \).
• Integrating \( N(x, y) \) with respect to \( y \).
• Combining terms to find the general solution \( \psi(x, y) = C \) where \( C \) is a constant.

For non-exact equations, as in this case, one might need to employ other methods, like an integrating factor, to find a workable solution.