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. \(t y^{2} y^{\prime}+2 y^{3}=1, \quad y(1)=-1\)

Short answer

Short Answer: The given differential equation is \(t y^2 y' + 2y^3 = 1\), and it is not exact. After finding the integrating factor \(\mu(t) = t^5\), we multiply the differential equation by \(\mu(t)\) to obtain an exact equation \((t^6 y^2) y' + 2t^5y^3 = t^5\). The solution to this exact equation is \(t^5y^4 = \frac{1}{7}t^7y^2\). However, applying the initial condition \(y(1) = -1\) leads to a contradiction, which means there is no explicit solution for the given problem under the given initial condition.

Step-by-step solution

Checking for exactness and finding integrating factor (\(\mu\))

First, let's identify the given differential equation, which is \(t y^2 y' + 2y^3 = 1\). To check whether this equation is exact or not, let's identify \(M(t, y) = 2y^3\) and \(N(t, y) = ty^2\). We need to find the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\), and check for exactness: \(\frac{\partial M}{\partial y} = 6y^2\) \(\frac{\partial N}{\partial t} = y^2\) Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}\), the given equation is not exact. So, we need to find an integrating factor, \(\mu\). According to the ideas of Exercise 22, if \(\mu\) depends only on \(t\), then \(\frac{\mu_t}{\mu} = \frac{\partial M/\partial y - \partial N/\partial t}{N}\). Let's find \(\mu_t\): \(\frac{\mu'_t}{\mu} = \frac{6y^2 - y^2}{ty^2} = \frac{5}{t}\) Integrating both sides with respect to \(t\): \(\int \frac{\mu'_t}{\mu} dt = \int \frac{5}{t} dt\) \(\ln |\mu| = 5 \ln |t| + C\) Now, we find the integrating factor \(\mu\): \(\mu(t) = t^5\)

Multiply by integrating factor and find exact equation

Let's multiply the given differential equation by the integrating factor, \(\mu(t) = t^5\), and obtain the exact equation: \(t^5(t y^2 y' + 2y^3) = t^5\) \((t^6 y^2) y' + 2t^5y^3 = t^5\) Now, check for exactness by finding partial derivatives of new \(M(t, y) = 2t^5y^3\) and \(N(t, y) = t^6y^2\): \(\frac{\partial M}{\partial y} = 6t^5y^2\) \(\frac{\partial N}{\partial t} = 6t^5y^2\) Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\), the equation is now exact.

Solve the exact differential equation and apply initial condition

The exact differential equation is \((t^6 y^2) y' + 2t^5y^3 = t^5\). To solve the exact equation, we integrate \(M(t, y)\) with respect to \(y\), and \(N(t, y)\) with respect to \(t\): \(\int M(t, y) dy = \int 2t^5y^3 dy = t^5y^4 + c_1(t)\) \(\int N(t, y) dt = \int t^6y^2 dt = \frac{1}{7}t^7y^2 + c_2(y)\) Comparing these two expressions, we can infer that \(c_1(t) = 0\) and \(c_2(y) = 0\). Thus, the solution to the exact equation is: \(t^5y^4 = \frac{1}{7}t^7y^2\) Now, applying the initial condition \(y(1) = -1\), we get: \(1(-1)^4 = \frac{1}{7}(1)^7(-1)^2\) \(1 = \frac{1}{7}\) This is a contradiction, which means that there is no explicit solution for the given problem under the given initial condition.

Key concepts

Integrating Factor

Exact differential equations can sometimes be difficult to solve when they aren't in an exact form. That's where an integrating factor can save the day. An integrating factor is a function that, when multiplied by a non-exact differential equation, transforms it into an exact one. This allows us to apply techniques for solving exact equations.
To find an integrating factor, often denoted as \(\mu(t)\), you must look for a function that, when multiplied by the given differential equation, makes the equation satisfy the exactness condition. In this scenario, we used the discrepancy between the partial derivatives of the components of the differential equation to find \(\mu\). The critical formula to remember is:

• \(\frac{\mu_t}{\mu} = \frac{\partial M/\partial y - \partial N/\partial t}{N}\)

Once you determine the form of the integrating factor by integrating with respect to \(t\), you get a new form of the differential equation. With the integrating factor \(\mu(t) = t^5\), it simplified our equation to an exact form so we could proceed to solve it.

Partial Derivatives

In mathematics, particularly calculus, partial derivatives are a way to show how a function changes as one of its dimensions changes, while the other dimensions are held constant. When dealing with differential equations, partial derivatives are essential in checking the exactness of an equation.
For an equation to be considered exact, the partial derivatives of its terms must satisfy the condition:

• \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\)

Here, \(M(t, y)\) and \(N(t, y)\) represent two different components of the differential equation. In our example, by calculating the partial derivative \(\frac{\partial M}{\partial y} = 6y^2\) and \(\frac{\partial N}{\partial t} = y^2\), it's clear that the two are not equal, indicating the original differential equation is not exact. Therefore, using partial derivatives is a step toward identifying the need for an integrating factor.

Initial Condition

An initial condition in differential equations is like the starting point of your favorite hiking trail; it tells you where to begin the solution process. An initial condition is often given as a specific value of the function or its derivatives at a particular point. This allows you to find a unique solution to a differential equation rather than a whole family of solutions.
In the problem we are considering, the initial condition is given as \(y(1) = -1\). When solving exact equations, after integrating and forming the general solution of the equation, you apply the initial condition to solve for any constants that arise during the integration process.
Unfortunately, in our exercise, applying the initial condition revealed a contradiction, \(1 = \frac{1}{7}\). This tells us that no solution satisfies both the differential equation and the initial condition. Thus, understanding and correctly applying initial conditions is crucial in ensuring that the solutions obtained are feasible.