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. \(\left(t^{2} y^{2}+1\right) y^{\prime}+t y^{3}=0, \quad y(0)=1, \quad \mu(t, y)=y^{-1}\)

Short answer

Short Answer: To show that the given non-exact differential equation is not exact, we calculate the partial derivatives of the functions M and N and show that they are not equal. After multiplying the equation by the given integrating factor μ(t,y) = y^(-1), we obtain a new equation and show that this new equation is exact by proving that the new partial derivatives are equal. We solve this exact equation, and the resulting solution cannot be simplified further. We obtain an implicit solution: \(y = e^{-\frac{1}{2} t^2 y^2}\). Although we cannot find an explicit solution, numerical solutions can be obtained if necessary.

Step-by-step solution

Determine the given differential equation is not exact

To show that the given differential equation is not exact, we first write it in the form \((M + N y')=0\), where \(M = t y^{3}\), \(N = t^{2} y^{2} + 1\), and \(y' = \frac{dy}{dt}\). Then, we calculate the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\). If these are equal, the differential equation is exact. If they are not equal, the differential equation is not exact.

Calculate the partial derivatives of M and N

Calculate the partial derivative of \(M\) with respect to \(y\): \(\frac{\partial M}{\partial y} = \frac{\partial (t y^3)}{\partial y} = 3t y^2\) Calculate the partial derivative of \(N\) with respect to \(t\): \(\frac{\partial N}{\partial t} = \frac{\partial (t^2 y^2 + 1)}{\partial t} = 2t y^2\) These partial derivatives are not equal, so the given differential equation is not exact.

Multiply the equation by the integrating factor \(\mu(t,y)\)

The integrating factor given is \(\mu (t,y) = y^{-1}\). So, we multiply the equation by \(y^{-1}\). \((M\mu) + (N\mu)y' = 0\), where \(M\mu = t y^{2}\) and \(N\mu = t^{2} y + y^{-1}\).

Determine the new equation is exact

Now, we must check if the modified equation is exact. We calculate the new partial derivatives \(\frac{\partial M\mu}{\partial y}\) and \(\frac{\partial N\mu}{\partial t}\) for the modified equation. Calculate the partial derivative of \(M\mu\) with respect to \(y\): \(\frac{\partial M\mu}{\partial y} = \frac{\partial (ty^2)}{\partial y} = 2ty\) Calculate the partial derivative of \(N\mu\) with respect to \(t\): \(\frac{\partial N\mu}{\partial t} = \frac{\partial (t^2 y + y^{-1})}{\partial t} = 2ty\) Since these partial derivatives are equal, the modified equation is exact.

Solve the exact equation

We can now solve the exact equation with the form \((M\mu) + (N\mu) y' = 0\) by integrating both terms with respect to their respective variables. Integrate \(M\mu = t y^2\) with respect to \(t\): \(\int M\mu dt = \int ty^2 dt = \frac{1}{2}t^2 y^2 + g(y)\) Integrate \(N\mu = t^2 y + y^{-1}\) with respect to \(y\): \(\int N\mu dy = \int (t^2 y + y^{-1}) dy = \frac{1}{2} t^2 y^2 + \ln |y| + h(t)\) Comparing the results, we find that \(g(y) = \ln |y|\) and \(h(t) = 0+ C\). Thus: \(\frac{1}{2} t^2 y^2 + \ln |y|=C\)

Solve for the initial condition

We are given the initial condition \(y(0) = 1\). Plug this into the equation to solve for C. \(\frac{1}{2} (0)^2 (1)^2 + \ln |1|=C\) \(C=0\) Thus, the equation becomes: \(\frac{1}{2} t^2 y^2 + \ln |y|=0\)

Solve for y(t)

We now solve for \(y(t)\) by isolating it from the equation: \(\ln |y| = -\frac{1}{2} t^2 y^2\) \(y = e^{-\frac{1}{2} t^2 y^2}\) This implicit solution cannot be further simplified. However, the numerical solution can be obtained through numerical methods if necessary.

Key concepts

Integrating Factor

In solving differential equations, an integrating factor is a function that simplifies the equation into a solvable form. It's particularly useful when the equation is not exact, meaning the conditions for an exact equation are not met. An integrating factor can often be found by recognizing a pattern or using a formula related to the differential equation's coefficients. When multiplied by the entire equation, it turns a non-exact differential equation into an exact one. This transformation allows us to integrate the equation and find a solution. For example, in the given exercise, the function \(\mu(t, y)=y^{-1}\) is provided as the integrating factor. When we multiply the non-exact differential equation by \(y^{-1}\), it becomes exact, allowing us to proceed with finding a solution.

Partial Derivatives

The concept of partial derivatives is crucial when dealing with exact differential equations. A partial derivative represents the rate at which a function changes as one of its variables changes, with all other variables held constant. To determine if a differential equation \(M + N y' = 0\) is exact, you should check if the partial derivative of \(M\) with respect to \(y\) is equal to the partial derivative of \(N\) with respect to \(t\). If these partial derivatives are not equal, then the equation is not exact, which is the case in our exercise. However, applying the integrating factor can transform it into an exact equation, as the new partial derivatives of the modified terms will then match, indicating that the equation can be solved through integration.

Initial Value Problem

An initial value problem involves a differential equation along with an initial condition, which specifies the value of the unknown function at a particular point. This information is pivotal as it allows us to find a unique solution to the differential equation, which could otherwise have an infinite number of solutions. In our textbook exercise, the initial condition \(y(0) = 1\) is provided. Once the general solution is found, this condition is used to determine a specific value for the constant of integration \(C\), leading to a particular solution that satisfies both the differential equation and the initial condition.

Explicit Solution

An explicit solution of a differential equation is one where the dependent variable is expressed solely in terms of the independent variable and constants. In other words, it's a straightforward expression such as \(y = f(t)\) without involving \(y\) implicitly. While such solutions are often more desirable and easier to interpret, they are not always obtainable. In the given exercise, although we seek an explicit solution, it's not always possible, especially if the relationship between the variables is complex, and that’s when an implicit solution becomes acceptable.

Implicit Solution

Contrary to an explicit solution, an implicit solution to a differential equation is one where the relationship between the dependent and independent variables is stated as an equation that is not explicitly solved for the dependent variable. For example, an equation of the form \(F(t, y) = 0\) represents an implicit solution. In the case of our exercise, the resulting solution \(\frac{1}{2} t^2 y^2 + \ln |y|=0\) cannot be algebraically solved for \(y\) in terms of \(t\), so it remains implicit. Implicit solutions are still valid and can offer valuable insights or be approached numerically if a precise function form is needed.