Problem 32 - Differential Equations and Linear Algebra

Question

Show that the given relation defines an implicit solution to the given differential equation, where \(c\) is an arbitrary constant. \(x^{2} y^{2}-\sin x=c, \quad y^{\prime}=\frac{\cos x-2 x y^{2}}{2 x^{2} y} .\) Determine the explicit solution that satisfies \(y(\pi)=1 / \pi\).

Step-by-Step Solution

Answer

The given relation, \(x^2y^2 - \sin x = c\), is an implicit solution to the provided differential equation \(y' = \frac{\cos x - 2xy^2}{2x^2y}\) as shown in the step-by-step solution. Using the initial condition \(y(\pi) = \frac{1}{\pi}\), we determine that \(c = 1\). Therefore, the implicit solution is given by \(x^2y^2 - \sin x = 1\). To find the explicit solution, numerical methods are recommended, as an algebraic solution for \(y(x)\) is not easily obtained.

Step 1: Differentiate the given relation with respect to x

To show that the given relation defines an implicit solution to the given differential equation, we need to differentiate the given relation with respect to x and show that the resulting expression matches the given differential equation. Differentiating the given relation, \(x^2y^2 - \sin x=c\) with respect to x, we get: \[\frac{d}{dx}(x^2y^2) - \frac{d}{dx}(\sin x) = \frac{d}{dx}(c)\] Now, differentiate the terms on the left side: \[2x \cdot y^2 + x^2 \cdot 2yy' - \cos x = 0\] Rearrange the equation to get the expression for \(y'\): \[y' = \frac{\cos x - 2xy^2}{2x^2y}\] This matches the given differential equation, so the given relation does define an implicit solution.

Step 2: Determine the explicit solution

Now, we must determine the explicit solution that satisfies the given initial condition \(y(\pi) = 1/ \pi\). Substitute \(x=\pi\) and \(y=1/\pi\) into the given relation, \(x^2y^2 - \sin x=c\): \[(\pi)^2(1/\pi)^2 - \sin \pi = c\] Solve for \(c\): \[c = (\pi)^2(1/\pi)^2 - 0 = 1\] So, the implicit solution is given by: \[x^2y^2 -\sin x = 1\] It is quite complex to determine the explicit formula for y(x). We suggest that you use numerical methods to solve for y(x).

Key Concepts

Differential Equation

A differential equation is a mathematical expression that relates a function with its derivatives. In many real-world applications, from engineering to economics, these equations are used to describe the behavior of dynamic systems. The distinction between ordinary differential equations (ODEs) and partial differential equations (PDEs) is based on whether they involve ordinary derivatives or partial derivatives, respectively.

For instance, the equation presented in the exercise, \(y^\prime=(\frac{\cos x-2 x y^{2}}{2 x^{2} y})\), is an ODE because it includes the derivative of the function \(y\) with respect to a single variable \(x\). ODEs can range from simple, linear, or first-order equations to complex, nonlinear, or higher-order ones.

Solving Differential Equations

Solving a differential equation involves finding a function or a set of functions that satisfy the relationship outlined in the equation. Solutions to differential equations can be either explicit or implicit. An explicit solution provides the function directly in the form \(y=f(x)\), whereas an implicit solution gives a relationship between \(y\) and \(x\) without explicitly solving for \(y\). The process of finding a solution varies based on the complexity of the equation and can involve techniques such as separation of variables, integrating factors, or the use of numerical methods when an analytical solution is complex or unavailable, as suggested for the explicit solution of the equation in the exercise.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function that is not isolated on one side of the equation. In the exercise, implicit differentiation is utilized to show that the given relation \(x^{2} y^{2}-\sin x=c\) is indeed the implicit solution of the differential equation. By treating \(y\) as a function of \(x\) and differentiating both sides of the equation with respect to \(x\), we include \(y'\), the derivative of \(y\) with respect to \(x\), in our equation. This technique allows us to find the relationship between \(y'\) and the other variables and constants in the equation, facilitating the verification of the solution.

Initial Value Problem

An initial value problem (IVP) is a specific type of differential equation paired with initial conditions that specify the values of the unknown function and its derivatives at a certain point. The purpose of the IVP is to find a unique solution that not only satisfies the differential equation but also the provided initial conditions. In the given exercise, once the implicit solution to the differential equation is verified using implicit differentiation, the next step is to satisfy the initial condition \(y(\pi)=1/\pi\). This condition helps determine the arbitrary constant \(c\) involved in the solution, which is integral to solving many types of differential equations, especially when a unique solution is required. By substituting the initial values into the implicit solution, students can isolate and solve for the unknown constant and work towards an explicit expression of the solution if possible.