Problem 28 - Advanced Engineering Mathematics

Question

Find the value of \(k\) so that the given differential equation is exact. $$ \left(6 x y^{3}+\cos y\right) d x+\left(2 k x^{2} y^{2}-x \sin y\right) d y=0 $$

Step-by-Step Solution

Answer

The value of \( k \) is 4.5.

Step 1: Identify Components

First, identify the functions \( M(x, y) \) and \( N(x, y) \) from the differential equation \( M(x, y) \, dx + N(x, y) \, dy = 0 \). We have \( M(x, y) = 6xy^3 + \cos y \) and \( N(x, y) = 2kx^2y^2 - x\sin y \).

Step 2: Compute Partial Derivatives

Compute the partial derivative \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).\[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(6xy^3 + \cos y) = 18xy^2 - \sin y \]\[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2kx^2y^2 - x\sin y) = 4kxy^2 - \sin y \]

Step 3: Set Partial Derivatives Equal

For the differential equation to be exact, the cross partial derivatives must be equal: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]Substituting the derivatives we computed:\[ 18xy^2 - \sin y = 4kxy^2 - \sin y \]

Step 4: Solve for k

By canceling \( -\sin y \) from both sides, we get:\[ 18xy^2 = 4kxy^2 \]Divide both sides by \( xy^2 \) (and assuming \( xy^2 eq 0 \)):\[ 18 = 4k \]Solve for \( k \):\[ k = \frac{18}{4} = 4.5 \]

Key Concepts

Partial Derivatives

Partial derivatives are an essential tool in calculus, especially when dealing with functions of multiple variables. If you have a function like \( M(x, y) \) or \( N(x, y) \), the partial derivative with respect to one variable involves differentiating the function while assuming the other variables are constants. This allows us to understand how the function changes as we vary only one variable.For example, if you need to find the partial derivative \( \frac{\partial M}{\partial y} \) of \( M(x, y) = 6xy^3 + \cos y \), you would treat \( x \) as a constant. Here’s a simple breakdown of the process:

• The derivative of \( 6xy^3 \) with respect to \( y \) is computed as \( 18xy^2 \), keeping \( x \) constant.
• The derivative of \( \cos y \) with respect to \( y \) is \( -\sin y \).

Thus, the partial derivative \( \frac{\partial M}{\partial y} = 18xy^2 - \sin y \). Remember, partial derivatives give us a snapshot of rate of change in one direction, simplifying the analysis of multivariable functions.

Cross Partial Derivatives

Cross partial derivatives are the set of partial derivatives taken in sequence involving different variables. In exact differential equations, the key property of cross partial derivatives is their equality under specific conditions.In our exercise, we deal with these cross derivatives to check if an equation is exact. For the equation to be exact, \( \frac{\partial M}{\partial y} \) should equal \( \frac{\partial N}{\partial x} \). These derivatives may seem complicated, but solving them step-by-step makes it easier.For instance, when you find \( \frac{\partial N}{\partial x} \), treating \( y \) as a constant works similarly as before:

• With \( N(x, y) = 2kx^2y^2 - x\sin y \), differentiating with respect to \( x \) gives \( 4kxy^2 - \sin y \).

Cross partial derivatives being equal (\( 18xy^2 - \sin y = 4kxy^2 - \sin y \)) ensures that the solution path doesn't depend on the route taken in the variable space, confirming exactness.

Exactness Condition

The exactness condition in differential equations is crucial for solvability without further transformation. It relates to ensuring the path independence of a function, meaning the final integrated value depends only on the start and end points, not on the path taken between them.For a given differential equation, say \( M(x, y) \, dx + N(x, y) \, dy = 0 \), the exactness condition is met if\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]This equality tells us that these differential elements can be integrated directly into a function \( \phi(x, y) \) that satisfies both components.In our case, discovering \( k \) such that the equation is exact (\( 18xy^2 = 4kxy^2 \)) involves precisely adjusting \( k \) to make these cross partials equal. Solving equations under such conditions streamlines to finding specific constants ensuring total compatibility, solidifying the status as exact.