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Determine whether the given differential equation is exact. $$2 x e^{y} d x+\left(3 y^{2}+x^{2} e^{y}\right) d y=0$$

Short Answer

Expert verified
The given differential equation is exact, as the partial derivatives match: \(\frac{\partial M}{\partial y} = 2xe^y\) and \(\frac{\partial N}{\partial x} = 2xe^y\).

Step by step solution

01

Compute the partial derivative of M with respect to y

We are given M(x, y) = 2xe^y, so we compute the partial derivative with respect to y: \(\frac{\partial M}{\partial y} = 2x\frac{d}{dy}(e^y) = 2xe^y\)
02

Compute the partial derivative of N with respect to x

We are given N(x, y) = 3y^2 + x^2e^y, so we compute the partial derivative with respect to x: \(\frac{\partial N}{\partial x} = 3\cdot0 + 2xe^y = 2xe^y\)
03

Compare the partial derivatives

Now we compare the partial derivatives we computed in steps 1 and 2: \(\frac{\partial M}{\partial y} = 2xe^y \\\) \(\frac{\partial N}{\partial x} = 2xe^y\) As both partial derivatives match, we can conclude that the given differential equation is exact.

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Key Concepts

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

Partial Derivatives
Understanding partial derivatives is crucial when dealing with multivariable functions, as they are indicative of a function's rate of change with respect to one variable while keeping other variables constant. In the context of exact differential equations, checking partial derivatives plays a significant role.

For example, when examining the function M(x, y) = 2xe^y, its partial derivative with respect to y, denoted as \( \frac{\partial M}{\partial y} \), is found by treating x as a constant and differentiating M solely with respect to y. This process is akin to ordinary differentiation, just applied to one variable at a time. Notably, the existence and equality of specific partial derivatives ensure that an equation is exact, serving as a test for exactness.
Differential Equation
A differential equation relates an unknown function to its derivatives. Understanding these equations is vital as many physical laws and relationships can be modeled using them. They span from simple separable equations to more complex ones, like the one in question:

\[2 x e^{y} dx + (3 y^{2} + x^{2} e^{y}) dy = 0\]
This particular equation is an example of an exact differential equation, where the solution involves finding a potential function whose partial derivatives correspond to the components of the equation. If a differential equation is accompanied by conditions known as boundary or initial conditions, it becomes a boundary-value or initial-value problem, respectively.
Exponential Function
The exponential function, often denoted as e to the power of a variable (like \(e^y\)), is notable for its constant rate of growth and its frequent occurrence in natural phenomena, such as population growth and radioactive decay.

In our differential equation, \(e^y\) indicates the equation's non-linearity and its variable-dependent growth rate. Exponential functions are special because their derivative is proportional to the function itself, making them invaluable in solving many types of differential equations. Moreover, the exponential function can also serve as an integrating factor, a tool we use to make some non-exact equations exact.
Integrating Factor
An integrating factor is a function used to transform a non-exact differential equation into an exact one, making it solvable through integration. The integrating factor is usually denoted as \(\mu(x, y)\) and is chosen such that it depends only on x or only on y to maintain simplicity in the computations.

When multiplied by the original differential equation, the integrating factor creates perfect differentials on both sides, which can then be integrated to find the solution. However, in the exercise we're considering, since both partial derivatives of \(M\) with respect to y and \(N\) with respect to x are equal, the equation is already exact, nullifying the need for an integrating factor in this instance.

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Most popular questions from this chapter

Solve $$ \sec ^{2} y \frac{d y}{d x}+\frac{1}{2 \sqrt{1+x}} \tan y=\frac{1}{2 \sqrt{1+x}} $$

(a) Show that the general solution to the differential equation $$ \frac{d y}{d x}=\frac{x+a y}{a x-y} $$ can be written in polar form as \(r=k e^{a \theta}\) (b) For the particular case when \(a=1 / 2,\) determine the solution satisfying the initial condition \(y(1)=1,\) and find the maximum \(x\) -interval on which this solution is valid. (Hint: When does the solution curve have a vertical tangent?) (c) \(\diamond\) On the same set of axes, sketch the spiral corresponding to your solution in (b), and the line \(y=x / 2\). Thus verify the \(x\) -interval obtained in (b) with the graph.

Consider the differential equation $$ y^{-1} y^{\prime}+p(x) \ln y=q(x) $$ where \(p(x)\) and \(q(x)\) are continuous functions on some interval \((a, b) .\) Show that the change of variables \(u=\ln y\) reduces Equation \((1.8 .26)\) to the linear differential equation $$ u^{\prime}+p(x) u=q(x) $$ and hence show that the general solution to Equation \((1.8 .26)\) is $$ y(x)=\exp \left\\{I^{-1}\left[\int I(x) q(x) d x+c\right]\right\\} $$ where $$ I=e^{\int p(x) d x} $$ and \(c\) is an arbitrary constant.

Solve the given differential equation. $$y^{\prime \prime}-2 x^{-1} y^{\prime}=18 x^{4}$$

Consider the phenomenon of exponential decay. This occurs when a population \(P(t)\) is governed by the differential equation $$\frac{d P}{d t}=k P$$ where \(k\) is a negative constant. A population of swans in a wildlife sanctuary is declining due to the presence of dangerous chemicals in the water. If the population of swans is experiencing exponential decay, and if there were 400 swans in the park at the beginning of the summer and 340 swans 30 days later, (a) how many swans are in the park 60 days after the start of summer? 100 days after the start of summer? (b) how long does it take for the population of swans to be cut in half? (This is known as the half-life of the population.)

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