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In Problems \(1-8,\) classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification. $$\left(y e^{x y}+2 x\right) d x+\left(x e^{x y}-2 y\right) d y=0$$

Short Answer

Expert verified
The given differential equation \( \left(y e^{x y}+2 x\right) d x+\left(x e^{x y}-2 y\right) d y = 0\) is exact but neither separable nor linear.

Step by step solution

01

Rewrite The Equation

Rewrite the original equation \( \left(y e^{x y}+2 x\right) d x+\left(x e^{x y}-2 y\right) d y = 0\) in terms of \(M(x, y)\) and \(N(x, y)\) as: \(M(x,y)dx + N(x,y)dy = 0\). Here, \(M(x, y) = ye^{xy}+2x\) and \(N(x, y) = xe^{xy}-2y\).
02

Check for Exactness

A differential equation of the form \(Mdx + Ndy = 0\) is exact if \(\partial M / \partial y = \partial N / \partial x\). Calculate the partial derivative \(\partial M / \partial y = xe^{xy} + yxe^{xy}\) and the partial derivative \(\partial N / \partial x = ye^{xy} + xye^{xy}\). Since \(\partial M / \partial y = \partial N / \partial x\), the given equation is exact.
03

Check for Separability and Linearity

The differential equation is not separable because you cannot separate the variables to different sides of the equation. Furthermore, it's not linear because it doesn't have the standard form for a linear differential equation \(a_1(x)y′′ + a_2(x)y′ + a_3(x)y = g(x)\).

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Key Concepts

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

Differential Equation Classification
When we encounter a differential equation, the first crucial step is to classify it. Differential equations can belong to various categories such as separable, linear, or exact. This classification helps us figure out the best way to solve them.

- **Separable**: An equation is separable if we can rearrange it so that one side of the equation involves only one variable and its derivatives, while the other side involves the other variable. Unfortunately, not all equations allow this separation.
- **Linear**: A linear differential equation has variables and derivatives raised only to the first power and follows a standard form like: \[ a_1(x)y'' + a_2(x)y' + a_3(x)y = g(x) \] Our equation does not fit this form, hence it is not linear.
- **Exact**: For a differential equation of form \( M(x, y)dx + N(x, y)dy = 0 \), it is exact if \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] If this condition is satisfied, as in our problem, it implies that the equation is indeed exact.

These classifications not only help in understanding the nature of the equation but also in applying the correct techniques for finding solutions.
Partial Derivatives
Partial derivatives are essential in the analysis of equations involving multiple variables. They allow us to focus on how one variable changes while keeping others constant. This is particularly important when dealing with multi-variable functions in differential equations.

In exact differential equations, we use partial derivatives to check the condition for exactness. Suppose we have - \( M(x, y) = ye^{xy} + 2x \) - \( N(x, y) = xe^{xy} - 2y \)
To check for exactness, compute:
  • \( \frac{\partial M}{\partial y} = xe^{xy} + yxe^{xy} \)
  • \( \frac{\partial N}{\partial x} = ye^{xy} + xye^{xy} \)
If these derivatives are equal, as in our exercise, it confirms that the differential equation is exact.

Understanding partial derivatives provides insight into how changes in one variable influence the entire function, which is particularly useful when verifying exactness in differential equations.
Non-linear Differential Equations
Non-linear differential equations are those where the variable, its function, or its derivatives appear with exponents other than one, are multiplied together, or are involved in functions like exponentials or trigonometric functions. They can be much more complex than linear equations, and they don't follow a specific general form like linear ones do.

In our problem, the presence of terms like \(e^{xy}\), which involve the product of the variables \(x\) and \(y\) in an exponential function, indicates non-linearity. This complexity makes finding general solutions tougher and often requires numerical or iterative methods, especially if the equation cannot be classified under exact, separable, or linear categories.

Handling non-linear differential equations usually demands sophisticated mathematical techniques or powerful computational tools since analytical solutions are rare. Understanding their structure deeply improves your problem-solving skills when tackling real-world scenarios that are often modeled by such non-linear equations.

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Most popular questions from this chapter

In Problems \(1-6,\) identify the equation as separable, linear, exact, or having an integrating factor that is a function of either \(x\) alone or \(y\) alone. $$\left(2 y^{2} x-y\right) d x+x d y=0$$

Riccati Equation. An equation of the form $$\frac{d y}{d x}=P(x) y^{2}+Q(x) y+R(x)$$ is called a generalized Riccati equation. (a) If one solution_say,$$u(x)-\text { of }(18)$$ is known, show that the substitution $$y=u+1 / v \text { reduces }(18)$$ to a linear equation in v. (b) Given that $$u(x)=x$$ is a solution to $$\frac{d y}{d x}=x^{3}(y-x)^{2}+\frac{y}{x}$$ use the result of part (a) to find all the other solutions to this equation. (The particular solution $$u(x)=x$$ can be found by inspection or by using a Taylor series method; see Section 8.1.)

Derive the following general formula for the solution to the Bernoulli equation (9): $$y = \left\\{ \begin{array} { l l } { \left[ \frac { ( 1 - n ) \int e ^ { ( 1 - n ) \int P ( x ) d x } Q ( x ) d x + C _ { 1 } ] ^ { 1 / ( 1 - n ) } } { e ^ { ( 1 - n ) \int P ( x ) d x } Q ( x ) d x + C _ { 1 } } \right] ^ { 1 / ( 1 - n ) } } & { \text { for } n \neq 1 } \\ { C _ { 2 } e ^ { \int [ Q ( x ) - P ( x ) ] d x } } & { \text { for } n = 1 } \end{array} \right.$$

Consider the initial value problem $$ \frac{d y}{d x}+\sqrt{1+\sin ^{2} x} y=x, \quad y(0)=2 $$. (a) Using definite integration, show that the integrating factor for the differential equation can be written as $$ \mu(x)=\exp \left(\int_{0}^{x} \sqrt{1+\sin ^{2} t} d t\right) $$ and that the solution to the initial value problem is $$ y(x)=\frac{1}{\mu(x)} \int_{0}^{x} \mu(s) s d s+\frac{2}{\mu(x)} $$. (b) Obtain an approximation to the solution at \( x=1 \) by using numerical integration (such as Simpson's rule, Appendix C) in a nested loop to estimate values of \( \mu(x) \) and, thereby, the value of $$ \int_{0}^{1} \mu(s) s d s $$. [Hint: First, use Simpson's rule to approximate \( \mu(x) \) at \( x=0.1,0.2, \dots, 1 \). Then use these values and apply Simpson's rule again to approximate \( \int_{0}^{1} \mu(s) s d s . ] \) (c) Use Euler's method (Section 1.4) to approximate the solution at \( x=1 \), with step sizes \( h=0.1 \) and 0.05. [A direct comparison of the merits of the two numerical schemes in parts (b) and (c) is very complicated, since it should take into account the number of functional evaluations in each algorithm as well as the inherent accuracies.]

Newton's Law of Cooling. According to Newton's law of cooling, if an object at temperature \( T \) is immersed in a medium having the constant temperature \( M \), then the rate of change of \( T \) is proportional to the difference of temperature \( M-T \). This gives the differential equation $$ d T / d t=k(M-T) $$ (a) Solve the differential equation for \( T \) (b) A thermometer reading \( 100^{\circ} \mathrm{F} \) is placed in a medium having a constant temperature of \( 70^{\circ} \mathrm{F} \). After 6 min, the thermometer reads \( 80^{\circ} \mathrm{F} \). What is the reading after 20 min? (Further applications of Newton's law of cooling appear in Section 3.3.)

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