Problem 18 - Advanced Engineering Mathematics

Question

Let \(M=2 y \sin x \cos x-y+2 y^{2} e^{x y^{2}}\) and \(N=-x+\sin ^{2} x+4 x y e^{x y^{2}}\) so that $$M_{y}=2 \sin x \cos x-1+4 x y^{3} e^{x y^{2}}+4 y e^{x y^{2}}=N_{x}$$ From \(f_{x}=2 y \sin x \cos x-y+2 y^{2} e^{x y^{2}}\) we obtain \(f=y \sin ^{2} x-x y+2 e^{x y^{2}}+h(y), h^{\prime}(y)=0,\) and \(h(y)=0\) solution is \(y \sin ^{2} x-x y+2 e^{x y^{2}}=c\)

Step-by-Step Solution

Answer

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

Step 1: Verify the Condition for Exactness

To determine if the given differential equation is exact, we need to verify that \( M_y = N_x \). This can be checked by differentiating \( M \) with respect to \( y \) and \( N \) with respect to \( x \), given: \( M_y = 2 \sin x \cos x - 1 + 4xy^3 e^{xy^2} + 4y e^{xy^2} \). Similarly, \( N_x = 2 \sin x \cos x - 1 + 4xy^3 e^{xy^2} + 4y e^{xy^2} \). Since \( M_y = N_x \), the differential equation is exact.

Step 2: Identify the Potential Function

Given the exact differential equation, identify the potential function \( f(x, y) \) such that \( f_x = M \) and \( f_y = N \). Start integrating \( M \) with respect to \( x \) to obtain the partial potential function: \( f = \int (2y \sin x \cos x - y + 2y^2 e^{xy^2})dx = y \sin^2 x - xy + 2 e^{xy^2} + h(y) \), where \( h(y) \) is an arbitrary function of \( y \).

Step 3: Differentiate and Solve for h(y)

Differentiate the potential function found in Step 2 with respect to \( y \) to check what \( h'(y) \) should be for the potential equation to satisfy \( f_y = N \). From \( N = -x + \sin^2 x + 4xy e^{xy^2} \), set the partial derivative equal to \( f_y \) and solve: \( f_y = \sin^2 x - x + 4xy e^{xy^2} + h'(y) = -x + \sin^2 x + 4xy e^{xy^2} \). This simplifies to \( h'(y) = 0 \), hence \( h(y) = 0 \).

Step 4: Write the General Solution

Now that \( h(y) \) is determined, the potential function \( f(x, y) = y \sin^2 x - xy + 2 e^{xy^2} \) can be used to write the solution to the differential equation. The general solution is \( f(x, y) = c \), or \( y \sin^2 x - xy + 2 e^{xy^2} = c \), where \( c \) is a constant.

Key Concepts

Potential Function

In the context of exact differential equations, a potential function acts as a bridge that connects the different partial derivatives and helps solve the equation. When you have an exact differential equation, there is a function, often denoted as \( f(x, y) \), whose total differential equals the original equation.The purpose of identifying a potential function is to simplify and solve the differential equation by reducing it to a recognizable form. For example, if given functions \( M \) and \( N \), constructing the potential function involves:

• Integrating \( M \) with respect to \( x \)
• Adding a function of \( y \), usually denoted as \( h(y) \), to incorporate terms related only to \( y \)

Once the potential function \( f(x, y) \) is determined, the solution to the differential equation can be expressed simply as \( f(x, y) = c \), where \( c \) is a constant.

Exactness Condition

The exactness condition is the criterion used to determine whether a differential equation is exact. It states that for a differential equation expressed as:\[ M(x, y)\,dx + N(x, y)\,dy = 0 \]to be exact, the partial derivative of \( M \) with respect to \( y \) must equal the partial derivative of \( N \) with respect to \( x \).Mathematically, this condition is represented as:\[ M_y = N_x \]Ensuring the equality of these partial derivatives means that the given differential equation can be represented as the total differential of some potential function \( f(x, y) \). This check assures that the equation can be solved using integration to find this potential function.

Differential Equation Solution

Once you have verified exactness and determined the potential function, you can find the differential equation's solution. The solution process involves using the potential function \( f(x, y) \) derived from previous steps.The general solution can be written as:

• \( f(x, y) = c \)

Here, \( c \) is an arbitrary constant. Substituting back the potential function provides a form that highlights the interdependent relationship of \( x \) and \( y \) according to the original differential equation.

Integration of Partial Derivatives

Integration of partial derivatives is a crucial part of solving exact differential equations. It allows you to reconstruct the potential function from the given differential equation components.To integrate partial derivatives properly, follow these general steps:

• Start by integrating \( M \) with respect to \( x \) to obtain a preliminary potential function. This step accounts for the \( x \)-dependent part of the potential function.
• Don't forget to add \( h(y) \), a function of \( y \), since integration in terms of \( x \) may miss terms involving only \( y \).
• Differentiate this integrated function with respect to \( y \) to determine any missing terms and compare with \( N \).
• Finally, solve for \( h(y) \) by setting the differential of the preliminary function equal to \( N \), ensuring that all terms align correctly.

Careful integration and understanding the relationship between the partial derivatives ensure that the potential function satisfies the original differential equation.