Problem 27 - Advanced Engineering Mathematics

Question

Find the value of \(k\) so that the given differential equation is exact. $$ \left(y^{3}+k x y^{4}-2 x\right) d x+\left(3 x y^{2}+20 x^{2} y^{3}\right) d y=0 $$

Step-by-Step Solution

Answer

The value of \(k\) is 10.

Step 1: Recognize the Form

The given differential equation is \((M(x,y)) dx + (N(x,y)) dy = 0\). Here, \(M(x, y) = y^3 + kxy^4 - 2x\) and \(N(x, y) = 3xy^2 + 20x^2y^3\). We need the equation to be exact, meaning \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

Step 2: Calculate \(\frac{\partial M}{\partial y}\)

Calculate the partial derivative of \(M(x, y)\) with respect to \(y\):\[\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y^3 + kxy^4 - 2x) = 3y^2 + 4kxy^3.\]

Step 3: Calculate \(\frac{\partial N}{\partial x}\)

Calculate the partial derivative of \(N(x, y)\) with respect to \(x\):\[\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(3xy^2 + 20x^2y^3) = 3y^2 + 40xy^3.\]

Step 4: Set Equations Equal

For exactness, set \(\frac{\partial M}{\partial y}\) equal to \(\frac{\partial N}{\partial x}\):\[3y^2 + 4kxy^3 = 3y^2 + 40xy^3.\] Simplify to find \(k\).

Step 5: Simplify and Solve for \(k\)

Cancel \(3y^2\) from both sides:\[4kxy^3 = 40xy^3.\] Divide by \(xy^3\) (assuming \(x eq 0\) and \(y eq 0\)):\[4k = 40.\] So, \(k = \frac{40}{4} = 10.\)

Key Concepts

Partial Derivatives

Partial derivatives are foundational in understanding changes in multivariable functions. Imagine you have a function, say, \( f(x, y) \), which depends on two variables, \( x \) and \( y \). A partial derivative allows you to analyze how \( f \) changes when one of the variables is varied, keeping the other constant.

In the exercise problem, we work with two functions, \( M(x, y) \) and \( N(x, y) \). To find out if the equation is exact, which means its solutions behave nicely and consistently, we need to compute partial derivatives, specifically:

• \( \frac{\partial M}{\partial y} \) - how \( M \) changes with \( y \)
• \( \frac{\partial N}{\partial x} \) - how \( N \) changes with \( x \)

Partial derivatives are computed like regular derivatives, but with the focus only on the variable of interest. For instance, when finding \( \frac{\partial M}{\partial y} \), treat \( x \) as a constant, and differentiate with respect to \( y \).

These derivatives help ensure that the change rates match, crucial for an equation's exactness.

Differential Equation Solving

Solving differential equations is about finding a function that describes a system. Equations like the one given in the exercise involve derivatives and express how one quantity changes relative to another.

An exact differential equation has a particular trait: there exists a function whose total differential equals the given equation. To verify an equation's exactness we equate partial derivatives computed from each term.

• The equation is of the form: \((M(x,y)) dx + (N(x,y)) dy = 0\)
• For exactness: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\)

Steps include recognizing the form, calculating partial derivatives, and confirming equality.

If the partial derivatives match, the equation is exact, allowing seamless function identification. In practice, this method finds solutions that trace dynamics of complex scenarios, like fluid flow or electrical circuits.

Calculus

Calculus is the mathematical study of change and motion, grounded in derivatives and integrals. It's indispensable in precisely analyzing variable relationships. Here, calculus helps solve the differential equation by implementing partial derivatives.

Key concepts include:

• Derivatives: Represent rate of change; used here to calculate partial derivatives.
• Integrals: Though not directly used, they encode the notion of "adding up" tiny pieces to find totals.
• Exact Equations: Involve matching conditions that make differential problems solvable.
  
  Calculus isn't just about static numbers; it's dynamic and extends every mathematical boundary. Each concept aids in exploring how one variable influences another's rate of change. This process allows us to understand deeper relationships and solve complex problems with prescribed solutions.