Question

Solving a First-Order Linear Differential Equation In Exercises \(51-58\) , solve the first-order differential equation by any appropriate method. $$ y^{\prime}=2 x \sqrt{1-y^{2}} $$

Short answer

The solution of the differential equation is \(y = \sin(x^2 + C) \)

Step-by-step solution

Recognize the equation type

Observe that the given equation has the form that allows it to be written in separable form. The given equation is \(y'=2x\sqrt{1-y^2}\), where \(y'\) represents the derivative of y with respect to x.

Arrange in separable form

Write the equation in separable form. To do this, separate the variables y to one side and x to the other like so: \( \frac{dy}{\sqrt{1-y^2}} = 2x\ dx\)

Integration of both sides

Integrate both sides of the equation with respect to their variables. This yields: \(\int \frac{dy}{\sqrt{1-y^2}} = \int 2x dx , the integrations yield arcsin(y) = x^2 + C\).

Solve for y

To solve for y, we need to remove the arcsin function. For this take sine of both sides as inverse sine and sine are inverse functions they cancel out. This gives: \(y = \sin(x^2 + C) \)

Key concepts

Separable Differential Equations

Separable differential equations are a subset of ordinary differential equations (ODEs) characterized by their ability to be arranged such that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the opposite side. To tackle these equations, we follow a few systematic steps.

Firstly, we identify if the given ODE can be classified as separable. In the case of the exercise, \(y'=2x\sqrt{1-y^2}\), this is achievable as the variables can be separated. Next, rearrange the equation so that each variable is with its differential. The process, demonstrated in Step 2 of the solution, involves algebraic manipulation to achieve the form \(\frac{dy}{\sqrt{1-y^2}} = 2x\ dx\).

Once in this form, we proceed by integrating both sides, treating each side independently. This method hinges upon our ability to handle the integrals that arise, which leads us to explore various integration techniques to find the solution.

Integrating Factors

While not directly used in solving the exercise provided, integrating factors are a potent tool for solving first-order linear differential equations that are not separable. An integrating factor is a function, typically denoted by \(\mu(x)\) or \(\mu(y)\), that when multiplied by the original equation, transforms it into an easily integrable form.

For the equation \(y' + p(x)y = q(x)\), an integrating factor \(\mu(x)\) can be found by the formula \(\mu(x) = e^{\int p(x)dx}\). The equation is then multiplied by this factor, resulting in an exact differential on the left-hand side, making the ODE solvable through direct integration. Again, while it's not utilized in the separable equation at hand, understanding integrating factors broadens a student's arsenal for attacking various types of differential equations.

Integration Techniques

When faced with finding an antiderivative or integrating a function, several techniques can be employed. The choice of method is highly dependent on the form of the function.

Simple functions may require basic methods like power rule or substitution, while more complex ones could necessitate integration by parts, partial fractions, or trigonometric substitutions. For Step 3 in the exercise, we deal with the integral \(\int \frac{dy}{\sqrt{1-y^2}}\), which falls into the category of an inverse trigonometric function-based integral.

The integral \(\int \frac{dy}{\sqrt{1-y^2}} = \arcsin(y) + C\), where \(C\) is the constant of integration, showcases the use of an inverse trigonometric function to simplify the expression. Students should be familiar with such functions and the associated integration results to solve such integrals effectively.

Inverse Trigonometric Functions

Inverse trigonometric functions find great use in calculus, especially in solving differential equations. They are the inverses of the sine, cosine, and tangent functions, and are commonly referred to as arcsine (\textbackslash arcsin), arccosine (\textbackslash arccos), and arctangent (\textbackslash arctan).

The relationship between inverse trigonometric functions and their respective primary functions is pivotal. For instance, if \(\arcsin(y) = x\), then it implies \(\sin(x) = y\). This inverse property is utilized in Step 4 of the exercise to isolate \(y\) from \(\arcsin(y) = x^2 + C\) by taking the sine of both sides, yielding \( y = \sin(x^2 + C)\).

Grasping these functions' concept, including their domains and ranges, allows students to untangle complex ODEs that can be transformed into a form that incorporates them. It is essential to recognize when and how to apply these inverse functions to successfully navigate the world of differential equations.