Question

In Exercises \(1-10,\) find the general solution to the exact differential equation. $$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{x}}$$

Short answer

The general solution of the given differential equation is \(y = \arcsin(x) - 2\sqrt{x} + C\)

Step-by-step solution

Write the equation in the standard form

Rewrite the equation \(dy/dx = 1/(\sqrt{1-x^{2}}) - 1/(\sqrt{x})\) to the standard form by isolating the derivative \(dy/dx\) on one side. It doesn't need to further modification in this case.

Separate the variables

To separate the variables, we multiply each side of the equation by \(dx\) to get \(dy = (1/(\sqrt{1-x^{2}}) - 1/(\sqrt{x}))dx\). This separation makes it easier to integrate later on.

Integrate both sides

Integrate each side with respect to the corresponding variable. \[ \int dy = \int (1/(\sqrt{1-x^{2}}) - 1/(\sqrt{x})) dx\] This leads to \[y = \arcsin(x) - 2\sqrt{x} + C \] where C is the constant of integration.

Key concepts

Separation of Variables

The method of separation of variables is a powerful technique used to solve ordinary differential equations, particularly when dealing with an exact differential equation like the one given in the exercise. In its essence, separation of variables involves rearranging the equation to isolate the two variables on different sides of the equation. Hence, each variable is only present on one side of the equation, separated by the equals sign.

This technique streamlines the process of finding a solution by allowing us to integrate the two sides independently. In the step-by-step solution provided, the differential equation \(dy/dx = 1/(\sqrt{1-x^{2}}) - 1/(\sqrt{x})\) is initially in a form that is suitable for separating the variables. Multiplying through by \(dx\) yields \(dy = (1/(\sqrt{1-x^{2}}) - 1/(\sqrt{x}))dx\), which clearly places the variable \(y\) and its differential \(dy\) on one side, and the variable \(x\) along with its differential \(dx\) on the other. The integration can then proceed smoothly, focusing on each side with its respective variable.

Integral Calculus

Integral calculus is a branch of mathematics that is concerned with the accumulation of quantities and the areas under and between curves. When we integrate a function, what we are essentially doing is summing the infinitesimal changes represented by the function's derivative, thus recovering the original function to which the derivative corresponds. In the process of solving a differential equation by separation of variables, we apply integral calculus to both sides of the separated equation to find the general solution.

In the exercise at hand, after separation of variables, we are left with the integration task: \[ \int dy = \int (1/(\sqrt{1-x^{2}}) - 1/(\sqrt{x})) dx\]. The integration with respect to \(y\) is straightforward, resulting in \(y\) alone. The integral with respect to \(x\), however, involves a difference of two terms that require some knowledge of integral calculus to evaluate. The first term leads to an inverse trigonometric function, \(\arcsin(x)\), and the second term results in a power function \(2\sqrt{x}\). The process confirms the power of integral calculus to handle complex expressions and emphasizes the importance of recognizing integrals that correspond to known basic functions or their combinations.

General Solution of Differential Equations

The general solution of a differential equation represents a family of solutions that includes all possible solutions to the equation. It typically involves an arbitrary constant because the differential equation by itself does not incorporate any particular initial conditions or boundary values. The arbitrary constant \(C\) represents the infinite number of specific solutions that could be derived by applying different initial conditions.

In the provided exercise, after performing the integration, we obtain \(y = \arcsin(x) - 2\sqrt{x} + C\), where \(C\) is the constant of integration. This expression is the general solution to the exact differential equation. It represents all possible antiderivatives that could fit with different specific scenarios or initial values. A particular solution can be obtained by substituting a known point \(x_{0}, y_{0}\) into this general solution to solve for \(C\). Understanding the concept of a general solution is crucial for grasping how differential equations can model a wide range of phenomena with varying initial conditions.