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Find an integrating factor and solve the given equation. $$ \left[4\left(x^{3} / y^{2}\right)+(3 / y)\right] d x+\left[3\left(x / y^{2}\right)+4 y\right] d y=0 $$

Short Answer

Expert verified
Question: Find the solution to the given exact differential equation: \(4\left(\frac{x^3}{y^2}\right) + \frac{3}{y} dx + 3\left(\frac{x}{y^2}\right) + 4y dy = 0\). Answer: The solution to the given exact differential equation is: \(4x^3e^{-3y - \frac{3}{y}} + 3xe^{-3y - \frac{3}{y}} + 3x^3ye^{-3y - \frac{3}{y}} - 4x^4e^{-3y - \frac{3}{y}} = C\).

Step by step solution

01

Identify Exact Differential Equation Components and Check if it is Exact

First, let's write the given differential equation in the standard form \(M(x,y)dx + N(x,y)dy = 0\), and identify the components \(M(x,y)\) and \(N(x,y)\): $$ M(x,y)=4\left(\frac{x^3}{y^2}\right) + \frac{3}{y}, \qquad N(x,y)=3\left(\frac{x}{y^2}\right) + 4y $$ Now, to check if the given equation is exact, we need to compute the partial derivatives of \(M\) and \(N\) with respect to \(y\) and \(x\), respectively. If these are equal, the equation is exact. $$ \frac{\partial M}{\partial y} = -8\frac{x^3}{y^3} - \frac{3}{y^2}, \qquad \frac{\partial N}{\partial x} = \frac{3}{y^2} $$ Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact.
02

Determine the Integrating Factor

To make this equation exact, we need to find an integrating factor \(\mu(x,y)\), which will be multiplied by both \(M(x,y)\) and \(N(x,y)\). We can find a suitable integrating factor by examining a few possible cases: \(\mu(x)\), \(\mu(y)\), or \(\mu(x,y)\). For simplicity, let us consider an integrating factor in the form of \(\mu(y)\). Our goal is to find a \(\mu(y)\) function such that: $$ \frac{\partial}{\partial y} (\mu(y)M(x,y)) = \frac{\partial}{\partial x} (\mu(y)N(x,y)) $$ Upon substitution, we get: $$ \frac{\partial}{\partial y}\left(\mu(y)\left(4\frac{x^3}{y^2}+\frac{3}{y}\right)\right)=\frac{\partial}{\partial x} \left(\mu(y)\left(3\frac{x}{y^2}+4y\right)\right) $$ We solve for \(\mu(y)\) by dividing both side of the equation by \(\mu(y)M(x,y)\): $$ \frac{1}{\mu(y)}\left(-8\frac{x^3}{y^3}-\frac{6}{y^2}\right) = \frac{3}{y^2} $$ Simplify and find the general solution for \(\mu(y)\): $$ \int \frac{6-3y^2}{y^2} dy = \int d\ln(\mu(y)) $$ $$ \ln(\mu(y)) = -3y - \frac{3}{y} + C $$ The arbitrary constant \(C\) is not important for the integrating factor. Thus: $$ \mu(y) = e^{-3y - \frac{3}{y}} $$
03

Apply the Integrating Factor and Check Exactness

Multiply both sides of the given differential equation with the integrating factor \(\mu(y)= e^{-3y - \frac{3}{y}}\), to obtain: $$ e^{-3y - \frac{3}{y}}(4\frac{x^3}{y^2}+\frac{3}{y}) dx+ e^{-3y - \frac{3}{y}}(3\frac{x}{y^2}+4y) dy=0 $$ Now, let's compute the partial derivatives of the new \(M(x,y)\) and \(N(x,y)\): $$ \frac{\partial}{\partial y} (e^{-3y - \frac{3}{y}}(4\frac{x^3}{y^2}+\frac{3}{y})) = \frac{\partial}{\partial x} (e^{-3y - \frac{3}{y}}(3\frac{x}{y^2}+4y)) $$ By verifying that the equation is now exact, we can move on to the next step to solve the equation.
04

Solve the Exact Equation

Now that we have an exact equation, we can find the function \(\Psi(x,y)\). To do this, we integrate the modified \(M(x,y)\) with respect to \(x\) and the modified \(N(x,y)\) with respect to \(y\), and then equate the results: $$ \int e^{-3y - \frac{3}{y}}(4\frac{x^3}{y^2}+\frac{3}{y}) dx = \int e^{-3y - \frac{3}{y}}(3\frac{x}{y^2}+4y) dy $$ Upon integrating both sides, we get: $$ 4x^3e^{-3y - \frac{3}{y}} + 3xe^{-3y - \frac{3}{y}} = -3x^3ye^{-3y - \frac{3}{y}} + 4x^4e^{-3y - \frac{3}{y}} + C $$ Finally, we can rewrite the expression to get the final solution: $$ 4x^3e^{-3y - \frac{3}{y}} + 3xe^{-3y - \frac{3}{y}} + 3x^3ye^{-3y - \frac{3}{y}} - 4x^4e^{-3y - \frac{3}{y}} = C $$

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Key Concepts

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

Exact Differential Equations

Exact differential equations are a specific type of first-order equations that can be solved by finding a potential function whose gradients create the components of the differential equation. Mathematically, if we have an equation in the form:

\[ M(x,y)dx + N(x,y)dy = 0 \]

For the equation to be exact, there must be a function \( \Psi(x,y) \) such that:

  • \( M(x,y) = \frac{\partial \Psi}{\partial x} \)
  • \( N(x,y) = \frac{\partial \Psi}{\partial y} \)

However, not all differential equations initially appear to be exact. The condition for exactness is that the partial derivative of \( M \) with respect to \( y \) is equal to the partial derivative of \( N \) with respect to \( x \). If the equation is not exact initially, we seek an integrating factor to transform it into an exact equation. Understanding exact differential equations is crucial because they provide a straightforward method for finding solutions by simply integrating, effectively reducing a potentially complex problem to a set of simpler tasks.

Partial Derivatives

Partial derivatives are foundational to understanding and solving differential equations, especially exact differential equations. When we have a function of two or more variables, like \( f(x, y) \), the partial derivative is the derivative of the function with respect to one variable, while holding the other variables constant.

For instance, when we have:

\[ \frac{\partial M}{\partial y} \text{ and } \frac{\partial N}{\partial x} \]

These represent the rate of change of \( M \) in the direction of \( y \) and \( N \) in the direction of \( x \), respectively. In our differential equation context, the equality of these particular partial derivatives is the criterion for an equation to be considered exact, as it implies the existence of a potential function \( \Psi(x,y) \). The concept of partial derivatives allows us to handle multivariable functions and is a powerful tool in mathematical analysis, physics, engineering, and many other fields where such functions naturally arise.

Differential Equations

Differential Equations are equations involving derivatives of unknown functions. They are central to modeling a wide range of phenomena in science and engineering. A first-order differential equation involves first derivatives of a function and often appears as:

\[ f(x, y)dx + g(x, y)dy = 0 \]

This represents a relationship between rates of change in different quantities. The objective in solving a differential equation is to find the unknown function \( y(x) \) or \( x(y) \) that satisfies the given relationship. Techniques to solve differential equations vary greatly, ranging from separation of variables, exact differential equations, integrating factors to more complex methods for higher-order or non-linear equations. Understanding the nature of differential equations and how to approach solving them is fundamental in predictive modeling and analysis in real-world scenarios.

Integrating Factor Method

The integrating factor method is employed to solve non-exact differential equations by essentially transforming them into exact ones. This is achieved by multiplying the entire equation by an appropriately chosen function, known as the integrating factor \( \mu(x,y) \), which depends solely on one variable (either \( x \) or \( y \)) or in some cases, on both variables.

To illustrate, the integrating factor will satisfy the condition:

\[ \frac{\partial}{\partial y} (\mu(y)M(x,y)) = \frac{\partial}{\partial x} (\mu(y)N(x,y)) \]

Once we find \( \mu(y) \), we multiply it across to mold our equation into an exact form. While this process may appear simple, it often involves recognizing patterns and solving side equations to determine the correct form of \( \mu \). Applying the integrating factor is a clever mathematical trick that can simplify complex differential equations, making them more approachable for solving.

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Most popular questions from this chapter

Suppose that a certain population satisfies the initial value problem \(d y / d t=r(t) y-k, \quad y(0)=y_{0}\) where the growth rate \(r(t)\) is given by \(r(t)=(1+\sin t) / 5\) and \(k\) represents the rate of predation. $$ \begin{array}{l}{\text { (a) Supposs that } k=1 / 5 \text { . Plot } y \text { versust for several values of } y_{0} \text { between } 1 / 2 \text { and } 1 \text { . }} \\ {\text { (b) P. Stimate the critical initial population } y_{e} \text { below which the population will become }} \\ {\text { extinct. }} \\ {\text { (c) Choose other values of } k \text { and find the correponding } y_{i} \text { for each one. }} \\ {\text { (d) Use the data you have found in parts }(a) \text { and }(b) \text { to plot } y_{c} \text { versus } k \text { . }}\end{array} $$

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transform the given initial value problem into an equivalent problem with the initial point at the origin. $$ d y / d t=t^{2}+y^{2}, \quad y(1)=2 $$

(a) Solve the Gompertz equation $$ d y / d t=r y \ln (K / y) $$ subject to the initial condition \(y(0)=y_{0}\) (b) For the data given in Example 1 in the text \([ \leftr=0.71 \text { per year, } K=80.5 \times 10^{6} \mathrm{kg}\), \right. \(\left.y_{0} / K=0.25\right]\), use the Gompertz model to find the predicted value of \(y(2) .\) (c) For the same data as in part (b), use the Gompertz model to find the time \(\tau\) at which \(y(\tau)=0.75 K .\) Hint: You may wish to let \(u=\ln (y / K)\).

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