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Verify that the given differential equation is exact; then solve it. $$ \left(\frac{2 x}{y}-\frac{3 y^{2}}{x^{4}}\right) d x+\left(\frac{2 y}{x^{3}}-\frac{x^{2}}{y^{2}}+\frac{1}{\sqrt{y}}\right) d y=0 $$

Short Answer

Expert verified
The solution is \( \frac{x^2}{y} + \frac{y^2}{x^3} + 2\sqrt{y} = C \).

Step by step solution

01

Identify Functions M and N

Given the differential equation \( M(x, y) \, dx + N(x, y) \, dy = 0 \), identify \( M(x, y) \) and \( N(x, y) \) from the equation: \( M(x, y) = \frac{2x}{y} - \frac{3y^2}{x^4} \) and \( N(x, y) = \frac{2y}{x^3} - \frac{x^2}{y^2} + \frac{1}{\sqrt{y}} \).
02

Compute Partial Derivatives

Calculate the partial derivative of \( M \) with respect to \( y \), \( M_y \), and \( N \) with respect to \( x \), \( N_x \).\[ M_y = -\frac{2x}{y^2} - \frac{6y}{x^4} \]\[ N_x = -\frac{6y}{x^4} + \frac{2x}{y^2} \]
03

Check for Exactness

The differential equation is exact if \( M_y = N_x \). Substitute the expressions to verify:\[ M_y = -\frac{2x}{y^2} - \frac{6y}{x^4} \]\[ N_x = -\frac{6y}{x^4} + \frac{2x}{y^2} \]Both expressions are equivalent (since they sum to the same value), confirming exactness.
04

Solve Using Potential Function \( \psi(x, y) \)

Since the equation is exact, a potential function \( \psi(x, y) \) such that \( \frac{\partial \psi}{\partial x} = M(x, y) \) and \( \frac{\partial \psi}{\partial y} = N(x, y) \) can be found.Integrate \( M(x, y) \) with respect to \( x \):\[ \psi(x, y) = \int \left( \frac{2x}{y} - \frac{3y^2}{x^4} \right) \, dx = \frac{x^2}{y} + \frac{y^2}{x^3} + h(y) \]Differentiate \( \psi(x, y) \) with respect to \( y \) and set equal to \( N(x, y) \) to find \( h'(y) \):\[ \frac{-x^2}{y^2} + h'(y) = \frac{2y}{x^3} - \frac{x^2}{y^2} + \frac{1}{\sqrt{y}} \]\[ h'(y) = \frac{2y}{x^3} + \frac{1}{\sqrt{y}} \]Integrate to find \( h(y) \):\[ h(y) = \frac{y^2}{x^3} + 2\sqrt{y} \]
05

Write the Solution

The function \( \psi(x, y) \) incorporating \( h(y) \) is then:\[ \psi(x, y) = \frac{x^2}{y} + \frac{y^2}{x^3} + 2\sqrt{y} = C \]This represents the implicit solution to the differential equation.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a key concept when dealing with functions of multiple variables, like in exact differential equations. In simple terms, a partial derivative measures how a function changes as just one of the variables changes, while the other variables are held constant. This is essential when working with functions in two variables, such as in the equation given. For the given differential equation, we identified two functions, \( M(x, y) \) and \( N(x, y) \), which are functions of \( x \) and \( y \). To check if the equation is exact, we calculated the partial derivatives: \( M_y \), the partial derivative of \( M \) with respect to \( y \), and \( N_x \), the partial derivative of \( N \) with respect to \( x \). To summarize the process:
  • Perform partial differentiation of \( M \) with respect to \( y \) and \( N \) with respect to \( x \).
  • Check the equality \( M_y = N_x \) to ensure the differential equation is exact.
Confirming this equality implies the paths along which a function changes are coherent, allowing us to find a potential function \( \psi(x, y) \).
Potential Function
A potential function, denoted as \( \psi(x, y) \), serves as a critical component in solving exact differential equations. For a given exact differential equation, the potential function is a function whose gradient (vector of partial derivatives) matches \( M(x, y) \) and \( N(x, y) \). Specifically, \( \frac{\partial \psi}{\partial x} = M(x, y) \) and \( \frac{\partial \psi}{\partial y} = N(x, y) \).Upon confirming the equation's exactness, we begin by solving for the potential function. This involves integrating \( M(x, y) \) with respect to \( x \):\[ \int M(x, y) \, dx = \psi(x, y) \]In our case, this resulted in \( \psi(x, y) = \frac{x^2}{y} + \frac{y^2}{x^3} + h(y) \), where \( h(y) \) is an unknown function of \( y \).To fully determine \( \psi(x, y) \), we differentiate the expression concerning \( y \) and equate it with \( N(x, y) \) to solve for \( h(y) \). Once obtained, \( \psi(x, y) = C \) forms the implicit solution.
Implicit Solution
An implicit solution is a way to express the solution to an equation without isolating the dependent variable. In the context of differential equations, an implicit solution is particularly straightforward in describing relationships that are inherently more complex.For exact differential equations, once the potential function \( \psi(x, y) \) is identified and computed, an implicit solution is expressed as:\[ \psi(x, y) = C \]where \( C \) is a constant. This equation cannot usually be solved for one variable in terms of another explicitly but provides an efficient way to encapsulate all the solutions at once. Returning to the example, our implicit solution is given as \( \frac{x^2}{y} + \frac{y^2}{x^3} + 2\sqrt{y} = C \). This compact form uses the intersections of the contours of \( \psi(x, y) \) as solutions to the differential equation, representing an entire family of curves, each differing by the constant \( C \).
Integration Techniques
Integration techniques are essential skills in calculus, especially when solving differential equations.Exact differential equations rely heavily on the ability to integrate parts of the function correctly. In our example, we must integrate \( M(x, y) \) with respect to \( x \) to form part of the potential function.Some useful integration techniques include:
  • Integration by Substitution: Simplifies integrals by changing variables, making the function easier to integrate.
  • Integration by Parts: Useful when the integrand is a product of two functions. Helps to break down complex integrals.
  • Trigonometric Substitutions: Simplifies integrals containing square roots by substituting trigonometric identities.
For this problem, the initial integration involved simpler algebraic functions, allowing us to integrate directly. This straightforward technique allowed us to isolate parts of \( \psi(x, y) \) and eventually derive the function \( h(y) \) to complete the potential function. Knowing when and how to apply these techniques can facilitate solving even the most challenging integrals encountered in exact differential equations.

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Most popular questions from this chapter

Suppose that a falling hailstone with density \(\delta=1\) starts from rest with negligible radius \(r=0 .\) Thereafter its radius is \(r=k t\) ( \(k\) is a constant) as it grows by accretion during its fall. Use Newton's second law- according to which the net force \(F\) acting on a possibly variable mass \(m\) equals the time rate of change \(d p / d t\) of its momentum \(p=m v-\) to set up and solve the initial value problem $$ \frac{d}{d t}(m v)=m g, \quad v(0)=0 $$ where \(m\) is the variable mass of the hailstone, \(v=d y / d t\) is its velocity, and the positive \(y\) -axis points downward. Then show that \(d v / d t=g / 4\). Thus the hailstone falls as though it were under one-fourth the influence of gravity.

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$$ y^{\prime}=y+y^{3} $$

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