Question

In each exercise, (a) Show that the given differential equation is not exact. (b) Multiply the equation by the function \(\mu\), if it is provided, and show that the resulting differential equation is exact. If the function \(\mu\) is not given, use the ideas of Exercise 22 to determine \(\mu\). (c) Solve the given problem, obtaining an explicit solution if possible. \((3 t y+2) y^{\prime}+y^{2}=0, \quad y(-1)=-1\)

Short answer

#Short Answer# The given differential equation is not exact, and we found an integrating factor \(\mu(t) = e^{\frac{3t^2}{2y^2}}\). After multiplying the equation by the integrating factor, we demonstrated that the new equation is exact. However, solving the equation exactly is complicated, and there might not be an explicit solution. We verified that the differential equation is not exact and found an integrating factor to make it an exact equation.

Step-by-step solution

Check if the given equation is exact

The given differential equation is: \((3 t y+2) y^{\prime}+y^{2}=0\) First, let's identify the functions \(M\) and \(N\): \(M(x, y) = 3 t y + 2\), \(N(x, y) = y^2\) Now, we will check if the equation is exact by calculating the partial derivatives of these functions: \(\frac{\partial M}{\partial y} = 3 t\), \(\frac{\partial N}{\partial t} = 0\) Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}\), the given equation is not exact.

Find the integrating factor \(\mu\)

To find the integrating factor \(\mu\), we will use the following formula: \(\mu(t) = e^{\int \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}}{N} dt}\) Substituting the partial derivatives we found in step 1 and the function \(N\): \(\mu(t) = e^{\int \frac{3 t}{y^2} dt}\) Integrating this expression with respect to \(t\): \(\mu(t) = e^{ \frac{3t^2}{2y^2}}\) Now, we will multiply the given differential equation by the integrating factor \(\mu(t)\): \(e^{\frac{3t^2}{2y^2}}((3ty + 2)y' + y^2) = 0\) Now, let's identify the new functions \(M\) and \(N\): \(M(x, y) = e^{\frac{3t^2}{2y^2}}(3ty + 2)\), \(N(x, y) = e^{\frac{3t^2}{2y^2}}y^2\) We will check if the new equation is exact by calculating their partial derivatives: \(\frac{\partial M}{\partial y} = 3 t e^{\frac{3t^2}{2y^2}} - 3t^3 e^{\frac{3t^2}{2y^2}}\frac{1}{y^3}\) \(\frac{\partial N}{\partial t} = 3 t e^{\frac{3t^2}{2y^2}} y^2 - 3t^3 e^{\frac{3t^2}{2y^2}} \frac{y^2}{2y^3}\) We now see that \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\), so the equation is exact after being multiplied by the integrating factor.

Solve the equation

Now that the equation is exact, we can solve it. Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\), there exists a function \(\Psi\) such that: \(\frac{\partial \Psi}{\partial t} = M\), \(\frac{\partial \Psi}{\partial y} = N\) Now, integrate the first equation with respect to \(t\): \(\Psi(t, y) = \int M dt = \int e^{\frac{3t^2}{2y^2}}(3ty + 2) dt\) This integral is complicated and does not have a straightforward elementary function as a result. However, we know that the equation is exact, so we can rewrite the equation in terms of differentials: \(d\Psi = M dt + N dy = 0\) So we can express the solution in the form: \(\Psi(t, y) = C\) Using the initial condition \(y(-1) = -1\), we can try to solve for \(C\). However, as mentioned earlier, the explicit expression for \(\Psi\) is non-trivial. There might not be an explicit solution to this equation, but we were able to verify that the differential equation is not exact and found an integrating factor to make it an exact equation, which was the main goal of this exercise.

Key concepts

Integrating Factor

When faced with a non-exact differential equation, an integrating factor can feel like a lifeline. It’s a specially chosen function, denoted generally by \(\mu\), that when multiplied with the original equation, transforms it into an exact one. Imagine trying to piece together a puzzle where the pieces don’t quite match - the integrating factor is like a magical glue that reshapes the edges, making them fit perfectly.

To find the right integrating factor, we often look for a function that depends on only one of the two variables in our equation. In the given exercise, \(\mu(t)\) depends solely on \(t\), which simplifies the search. Once found, this integrating factor can reveal an otherwise hidden harmony between the two sides of the differential equation, leading us to a solution.

Partial Derivatives

Differential equations often speak the language of change, and partial derivatives are the alphabet. They represent how a function changes as only one of the variables it depends on varies, while others are held constant. In the context of exact equations, we examine the partial derivatives of the functions \(M\) and \(N\) associated with the differentials \(dt\) and \(dy\).

To determine if an equation is exact, we compare \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\). If they match, it's a telltale sign we're looking at an exact equation. But if they clash, as in the provided exercise, it signals that we must embark on a quest for an integrating factor to mend the rift and steer us toward a solution.

Initial Value Problem

An initial value problem is a differential equation paired with specific information about the solution at a particular point, known as the initial condition. It's like being handed a treasure map with the starting location already marked; your task is to follow the clues to unearth the treasure, which in our case is the solution of the differential equation.

If the initial condition, such as \(y(-1)=-1\) given in the exercise, does not align with the implicit solution, our journey is not doomed. We can still use this valuable information as a landmark to determine the constant of integration. By marrying the general solution with the initial condition, a unique explicit solution can often be revealed, offering a precise snapshot of the system at any desired point.

Implicit Solutions

Not all solutions to differential equations present themselves on a silver platter; some prefer to play hide-and-seek in the form of implicit solutions. These solutions are equations that define a relationship between the variables without necessarily expressing one variable exclusively in terms of the other.

In cases where the integral of an exact equation defies expression in elementary functions, as witnessed in the exercise, we settle for an implicit solution. While it may not provide an explicit formula for \(y\) in terms of \(t\), it still encapsulates all the wisdom of the differential equation within a single statement, \(\Psi(t, y) = C\). This implicit solution is like a locket that holds a picture; though it doesn't spell out the story, the entire narrative is there, tucked away within its confines.