Question

Finding a Particular Solution Using Separation of Variables In Exercises \(15-24\) , find the particular solution that satisfies the initial condition. $$ y \sqrt{1-x^{2}} y^{\prime}-x \sqrt{1-y^{2}}=0 \quad y(0)=1 $$

Short answer

The particular solution of the given differential equation that satisfies the initial condition \(y(0) = 1\) is \(y = \sin\left(\frac{x^{2}}{2} + \frac{\pi}{2}\right)\).

Step-by-step solution

Rewriting the equation

First, the given differential equation \(y \sqrt{1-x^{2}} y^{\prime}-x \sqrt{1-y^{2}}=0\) needs to be rewritten to isolate the derivative \(y'\). Divide the equation by \(y \sqrt{1-x^{2}}\) to get \(y'= \frac{x \sqrt{1-y^{2}}}{y \sqrt{1-x^{2}}}.\)

Separation of variables

Next, isolate all the terms involving \(y\) and \(x\) on separate sides of the equation. This can be done by multiplying both sides by \(y \sqrt{1-x^{2}}\) on the right and dividing both sides by \(\sqrt{1-y^{2}}\) on the left to obtain \(\frac{y'}{\sqrt{1-y^{2}}} = \frac{x}{y \sqrt{1-x^{2}}}\).

Integration

After the variables have been separated, integrate both sides of the equation. You will get \( \int \frac{y'}{\sqrt{1-y^{2}}} dy = \int \frac{x}{y\sqrt{1-x^{2}}} dx\). Here, the left hand side can be recognized as the integral form of inverse sine function and the right hand side is the integral of \(x\) with respect to \(x\), which gives \(\arcsin(y) = \frac{x^{2}}{2} + C\), where \(C\) is the constant of integration.

Applying the initial condition

Apply the initial condition \(y(0) = 1\) into the integral equation \(\arcsin(y) = \frac{x^{2}}{2} + C\). When \(x = 0\), \(y = 1\). So, we get \(\arcsin(1) = \frac{0^{2}}{2} + C\), which simplifies to give \(C = \frac{\pi}{2}\). Therefore, our equation becomes \(\arcsin(y) = \frac{x^{2}}{2} + \frac{\pi}{2}\).

Final Solution

Finally, to find \(y(x)\), apply the sine function to both sides of the equation to retrieve \(y\), thereby arriving at a final solution of \(y = \sin\left(\frac{x^{2}}{2} + \frac{\pi}{2}\right)\).

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function to its derivatives. They are essential in modeling various physical phenomena and processes, where the rate of change is a crucial element to understand. In simple terms, a differential equation gives us a way to calculate not just where something is (like a position), but also how it's moving or changing (speed, acceleration, growth rate, etc.) at any point in time or space.

For example, the given exercise involves a differential equation that captures the relationship between the changes in two variables, namely the function value, which refers to the coordinate 'y', and its derivative with respect to 'x', which could be the rate at which 'y' changes as 'x' changes. The equation incorporates both algebraic expressions and derivatives, making it a classic case for application of the separation of variables technique.

Initial Condition

The initial condition in the context of differential equations provides a starting point that a solution has to satisfy. It's essentially a snapshot of the system at a specific time. This information is crucial because, without it, a differential equation might have infinitely many solutions.

In the exercise, the initial condition given is 'y(0) = 1'. This tells us that when 'x' is zero, the value of 'y' is one. Using this piece of information helps to narrow down the infinite possibilities to one particular solution—here, the constant of integration 'C' can be precisely determined when solving the integral equation.

Integral Calculus

Integral calculus is the branch of mathematics focused on sums and accumulation. While its counterpart, differential calculus, deals with rates of change, integral calculus is about the total value accumulated from these changes. It's like putting together lots of tiny pieces to get a whole picture.

Within our exercise, we used integral calculus after separating the variables. The integration of a function often requires recognizing patterns or simplifying the equation into a form that matches a known integral. The integral of the left side was recognized as an inverse trigonometric function, specifically the arcsine function. The right side was a more straightforward algebraic integral, leading to a relationship between 'x' and 'y' after the integration process.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse operations of the basic trigonometric functions like sine, cosine, and tangent. These functions allow you to determine the angle when given a ratio of sides in a right triangle, thereby 'undoing' the original trigonometric function.

In the context of our exercise, the inverse trigonometric function arises naturally from the integration process. To be specific, the integral of \(1/\sqrt{1-y^{2}}\) with respect to 'y' is the arcsine of 'y'. Recognizing this relationship is key to solving the differential equation. After finding the general solution involving arcsine, we can use the initial condition to find the particular solution, leading us to the final expression for 'y' in terms of 'x'.