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Chapter 2: Problem 11

Determine whether the given differential equation is exact. If it is exact, solve it. $$\left(y \ln y-e^{-x y}\right) d x+\left(\frac{1}{y}+x \ln y\right) d y=0$$

Short Answer

Expert verified
The differential equation is not exact, so it cannot be solved using this method.

Step by step solution

01

Identify M(x,y) and N(x,y)

In a differential equation of the form \(M(x, y)\,dx + N(x, y)\,dy = 0\), first identify \(M(x, y) = y \ln y - e^{-xy}\) and \(N(x, y) = \frac{1}{y} + x \ln y\).
02

Calculate Partial Derivatives

Compute the partial derivative \(\frac{\partial M}{\partial y}\). This gives \(\frac{\partial M}{\partial y} = \ln y + 1 + xe^{-xy}\). Next, calculate \(\frac{\partial N}{\partial x}\), which is \(\frac{\partial N}{\partial x} = \ln y\).
03

Check for Exactness

For the equation to be exact, \(\frac{\partial M}{\partial y}\) must equal \(\frac{\partial N}{\partial x}\). Here, \(\frac{\partial M}{\partial y} = \ln y + 1 + xe^{-xy}\) is not equal to \(\frac{\partial N}{\partial x} = \ln y\).
04

Conclusion

Since \(\frac{\partial M}{\partial y}\) does not equal \(\frac{\partial N}{\partial x}\), the given differential equation is not exact. Hence, it is not solvable by the method for exact equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives play a crucial role in the study of multivariable calculus. They allow us to investigate how a function changes as we vary one of its variables, keeping the others constant. For a function of two variables, say \( f(x, y) \), the partial derivative with respect to \( x \) measures the rate of change of \( f \) while treating \( y \) as a constant.
Here are some key points to understand about partial derivatives:
  • Partial derivatives are denoted as \( \frac{\partial f}{\partial x} \) or \( \frac{\partial f}{\partial y} \).
  • They are computed similarly to ordinary derivatives, but they focus on one variable at a time.
  • In our exercise, we computed \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) to check the exactness of a differential equation.
Understanding partial derivatives is essential in solving and analyzing differential equations, especially those involving more than one variable.
Exactness in Differential Equations
Exactness in differential equations refers to a property that makes them potentially solvable in a straightforward manner. A first-order differential equation in the form \( M(x, y)\,dx + N(x, y)\,dy = 0 \) is said to be exact if the mixed partial derivatives of \( M \) and \( N \) are equal, specifically \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
Here are the main points:
  • Exact differential equations often stem from potential functions or conservative fields in physics.
  • An exact equation means there exists a function \( \Psi(x, y) \) such that \( \Psi_x = M(x, y) \) and \( \Psi_y = N(x, y) \).
  • The condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) ensures \( \Psi(x, y) \) is well-defined, allowing us to find solutions by integration.
For the problem at hand, the differential equation was not exact, as shown by the inequality of these partial derivatives, indicating the need for alternative solution methods like integrating factors or transformations.
Solving Differential Equations
Solving differential equations requires identifying appropriate methods based on their characteristics. For exact equations, if identified, the solution involves finding a potential function. However, when a differential equation is not exact, as in the exercise provided, other approaches must be pursued.
Key strategies for solving differential equations include:
  • **For exact equations**, identify a function \( \Psi(x, y) \) that satisfies \( \Psi_x = M(x, y) \) and \( \Psi_y = N(x, y) \).
  • **Using integrating factors** can sometimes turn a non-exact equation into an exact one.
  • **Transformations** or substitutions may simplify the equation, making it easier to solve.
  • **Numerical methods** can be considered for difficult equations not solvable analytically.
In our exercise, the differential equation was determined to be non-exact. This means the direct method using potential functions does not apply, necessitating other strategies to explore a possible solution.

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Most popular questions from this chapter

Use a technique of integration or a substitution to find an explicit solution of the given differential equation or initial value problem. $$\frac{d y}{d x}=\frac{1}{1+\sin x}$$

Find an explicit solution of the given initial-value problem. Determine the exact interval \(I\) of definition by analytical methods. Use a graphing utility to plot the graph of the solution. $$(2 y-2) \frac{d y}{d x}=3 x^{2}+4 x+2, \quad y(1)=-2$$

Solve the given differential equation by using an appropriate substitution. $$3\left(1+t^{2}\right) \frac{d y}{d t}=2 t y\left(y^{3}-1\right)$$

Solve the given initial-value problem. $$y^{1 / 2} \frac{d y}{d x}+y^{3 / 2}=1, \quad y(0)=4$$

Solve the given differential equation by finding, as in Example \(4,\) an appropriate integrating factor. $$y(x+y+1) d x+(x+2 y) d y=0$$
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