Question

Find the value of \(b\) for which the given equation is exact and then solve it using that value of \(b\). $$ \left(y e^{2 x y}+x\right) d x+b x e^{2 x y} d y=0 $$

Short answer

Question: For the exact differential equation, find the value of \(b\) and solve the equation for that value: $$ \left(ye^{2xy} + x\right) dx + bx e^{2xy} dy = 0 $$ Answer: The value of \(b\) for which the given equation is exact is \(b = 1\). The solution to the exact differential equation is: $$ \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} = C $$

Step-by-step solution

Determine the value of b for which the given equation is exact

To find the value of \(b\), we need to check for the exactness of the given equation. For a differential equation of the form \(M(x, y) dx + N(x, y) dy = 0\), the necessary condition for exactness is: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Let's find the partial derivatives for our given equation: $$ M(x,y) = ye^{2xy} + x \\ N(x,y) = bx e^{2xy} $$ Therefore, $$ \frac{\partial M}{\partial y} = \frac{\partial (ye^{2xy} + x)}{\partial y} = e^{2xy} + 2xye^{2xy} \\ \frac{\partial N}{\partial x} = \frac{\partial (bx e^{2xy})}{\partial x} = b e^{2xy} + 2bxy e^{2xy} $$ Now, equate \(\frac{\partial M}{\partial y}\) with \(\frac{\partial N}{\partial x}\): $$ e^{2xy} + 2xye^{2xy} = b e^{2xy} + 2bxy e^{2xy} $$ Since \(e^{2xy}\) doesn't equal zero, we can divide both sides of the equation by it: $$ 1 + 2xy = b + 2bxy $$ We can see that \(1 = b\) and \(2xy = 2xy\). Therefore, the value of \(b\) for which the given equation is exact is \(b = 1\).

Solve the exact differential equation using the value of b

Now that we have found the value of \(b\), we can rewrite our given exact differential equation as: $$ \left(ye^{2xy} + x\right) dx + xe^{2xy} dy = 0 $$ As the equation is exact, there exists a function \(\psi(x, y)\) such that \(\frac{\partial \psi}{\partial x} = M(x, y)\) and \(\frac{\partial \psi}{\partial y} = N(x, y)\). Let's find \(\psi(x, y)\) using this condition. Integrate \(\frac{\partial \psi}{\partial x}\) with respect to \(x\): $$ \psi(x, y) = \int (ye^{2xy} + x) dx $$ This integration gives: $$ \psi(x, y) = \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} + g(y) $$ Now we need to find the function \(g(y)\). To do that, differentiate \(\psi(x, y)\) with respect to \(y\): $$ \frac{\partial \psi}{\partial y} = \frac{\partial (\frac{1}{2} x^2 + \frac{1}{2}e^{2xy} + g(y))}{\partial y} = xe^{2xy} + g'(y) $$ We know that \(\frac{\partial \psi}{\partial y} = N(x, y) = xe^{2xy}\). Comparing the expressions, we get \(g'(y)=0\) which implies that \(g(y)\) is a constant, let's say \(C\). Our final solution will be in the form \(\psi(x, y) = C\): $$ \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} + C = 0 $$ So, the solution to the given exact differential equation for \(b = 1\) is: $$ \frac{1}{2} x^2 + \frac{1}{2}e^{2xy} = C $$

Key concepts

Partial derivatives

Partial derivatives are crucial in the study of multivariable functions where they indicate how a function changes as one variable changes while the others remain constant. In the context of exact differential equations, partial derivatives help verify whether a differential equation is exact. An exact equation is one where the partial derivative of one function with respect to one variable equals the partial derivative of another function with respect to the corresponding variable.

For example, in the provided exercise, functions \(M(x, y)\) and \(N(x, y)\) illustrate this relationship. By computing \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\), you can establish the exactness by checking if these derivatives are equal.

Important properties of partial derivatives include:

• Linearity: Partial derivatives respect addition and scalar multiplication.
• Product rule: Similar to single-variable calculus, but applied to each variable while treating others as constants.
• When partial derivatives match under these scrutiny conditions, an exact differential equation is confirmed.

Integration of functions

To solve exact differential equations, integrating functions is essential. After verifying exactness through partial derivatives, integration allows us to reconstruct the potential function from its partial derivatives.

For the exact differential equation, we integrate \(\frac{\partial \psi}{\partial x} = M(x, y)\) with respect to \(x\) to find \(\psi(x, y)\). In the exercise, this is shown by integrating terms like \(ye^{2xy} + x\). The challenge often lies in dealing with complex expressions that result from multivariable integration.

Key integration techniques include:

• Recognizing common patterns such as exponential and polynomial functions for straightforward integration.
• Substitution techniques when dealing with composite expressions, especially in exponential terms like \(e^{2xy}\).
• Constant of integration that arises is often a function of the other variable(s), capturing aspects not covered initially.

These strategies aid in finding the cumulative function \(\psi\) that satisfies the conditions of the differential equation.

Boundary Value Problems

Boundary value problems (BVPs) involve finding a solution to differential equations subject to certain boundary conditions. Though the provided exercise mainly discusses exact equations, understanding BVPs is critical when dealing with differential equations in practical applications.

Typically, a BVP specifies the value of a solution at specific points. For example, finding \(\psi(x, y)\) must not only satisfy the exactness conditions but also meet specific initial or boundary conditions if they are given.

Some essential elements of BVPs include:

• Boundary conditions: Can be either Dirichlet (specifies exact values) or Neumann (specifies derivative values).
• Existence and uniqueness: Solutions may or may not exist depending on the boundary conditions set.
• Real-world applications: Often found in physics and engineering where values at surfaces or interfaces are known.

Although BVPs weren't explicitly involved in this exercise, knowing them is useful for tackling more complex situations where boundary conditions play a crucial role.

Solving differential equations

Solving differential equations takes us through a structured process where initial verification of equation characteristics, such as exactness, informs the solution approach. The exercise demonstrates solving an exact differential equation through evaluation, integration, and derivational checks.

Steps in solving involve:

• Verification of exactness: Ensures that the equation holds constant partial derivative conditions.
• Integration of terms to find the potential function \(\psi(x, y)\), where each term is evaluated within its context.
• Adjustment with constant functions or terms \(g(y)\) to reconcile integration scopes.
• Final solution: Formulated as a family of curves \(\psi(x, y) = C\), representing solution sets for varying conditions.

Through meticulous application of these steps, the solution achieves the derivational and conditional harmonization required to solve exact differential equations effectively.