Problem 9 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The differential equation is not exact because the exactness condition is not satisfied.

Step 1: Identify Components

The given problem can be written in the form \( M(x,y)\,dx + N(x,y)\,dy = 0 \) where \( M(x,y) = x - y^3 + y^2 \sin x \) and \( N(x,y) = 3xy^2 + 2y \cos x \). We need to check if this differential equation is exact.

Step 2: Check Exactness Condition

For a differential equation of the form \( M\,dx + N\,dy = 0 \) to be exact, the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) must hold. Calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).

Step 3: Calculate Partial Derivatives

Compute \( \frac{\partial M}{\partial y} = -3y^2 + 2y \sin x \). Compute \( \frac{\partial N}{\partial x} = 3y^2 - 2y \sin x \).

Step 4: Compare Partial Derivatives

Compare \( \frac{\partial M}{\partial y} = -3y^2 + 2y \sin x \) and \( \frac{\partial N}{\partial x} = 3y^2 - 2y \sin x \). They are not equal (\( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \)), so the differential equation is not exact.

Key Concepts

Understanding Partial Derivatives

To determine if a given differential equation is exact, we need to work with partial derivatives. A partial derivative is essentially the derivative of a function with respect to one variable while holding the other variables constant. This concept is particularly important in multivariable calculus.
In the context of differential equations, we have functions such as \( M(x, y) \) and \( N(x, y) \) where each can depend on multiple variables. When computing \( \frac{\partial M}{\partial y} \), we differentiate \( M \) with respect to \( y \) while treating \( x \) as a constant. Similarly, \( \frac{\partial N}{\partial x} \) is computed by differentiating \( N \) with respect to \( x \), keeping \( y \) constant.
These derivatives help us determine the behavior of the equation and check for exactness, which is crucial in solving these types of equations.

The Exactness Condition Explained

A differential equation is in the form \( M\,dx + N\,dy = 0 \), and is considered exact when the exactness condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) holds true. This means that the partial derivative of \( M \) with respect to \( y \) should equal the partial derivative of \( N \) with respect to \( x \).
The reasoning behind this condition is rooted in the theory of exact differential equations, where an exact equation implies that there exists a potential function \( \phi(x, y) \) such that \( \phi_x = M \) and \( \phi_y = N \). This implies that the differentials \( d\phi = M\,dx + N\,dy = 0 \) will equal zero due to the equality of the mixed second partial derivatives. Validating this condition is a way to verify if a differential equation is genuinely exact and can thus be integrated directly.

Solutions for Differential Equations

When dealing with differential equations, the ultimate goal is often to find a solution. Once you determine that a differential equation is exact, solving it involves finding a potential function \( \phi(x, y) \).
If \( M(x, y)\, dx + N(x, y)\, dy = 0 \) is exact, then there exists a function \( \phi(x, y) \) such that \( d\phi = 0 \). To find \( \phi(x, y) \), perform the following steps:

• Integrate \( M \) with respect to \( x \), while treating \( y \) as constant. This will provide you a function including \( y \) as an integration constant.
• Differentiate this result with respect to \( y \) and equate it with \( N \), allowing you to find any terms of \( \phi(x, y) \) that were functions of \( y \) alone.

By constructing \( \phi(x, y) \), you effectively solve the differential equation for all \( x \) and \( y \) that satisfy the conditions, revealing the solution as implicit relationships mathematical solutions demand.