Problem 27 - Advanced Engineering Mathematics

Question

Separating variables and integrating we obtain \\[\frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0 \quad \text { and } \quad \sin ^{-1} x-\sin ^{-1} y=c\\]. Setting \(x=0\) and \(y=\sqrt{3} / 2\) we obtain \(c=-\pi / 3 .\) Thus, an implicit solution of the initial-value problem is \(\sin ^{-1} x-\sin ^{-1} y=\pi / 3 .\) Solving for \(y\) and using an addition formula from trigonometry, we get \\[y=\sin \left(\sin ^{-1} x+\frac{\pi}{3}\right)=x \cos \frac{\pi}{3}+\sqrt{1-x^{2}} \sin \frac{\pi}{3}=\frac{x}{2}+\frac{\sqrt{3} \sqrt{1-x^{2}}}{2}\\].

Step-by-Step Solution

Answer

The solution for \( y \) in terms of \( x \) is \( y = \frac{x}{2} + \frac{\sqrt{3} \sqrt{1-x^2}}{2} \).

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0 \). This form suggests using the integration method by separating variables.

Step 2: Separate Variables and Integrate

Re-arrange the terms to separate variables: \( \frac{d x}{\sqrt{1-x^{2}}}=\frac{d y}{\sqrt{1-y^{2}}} \). Integrate both sides: \( \int \frac{d x}{\sqrt{1-x^{2}}} = \int \frac{d y}{\sqrt{1-y^{2}}} \). This gives \( \sin^{-1} x = \sin^{-1} y + C \).

Step 3: Solve for Constant Using Initial Conditions

Given initial conditions, set \( x = 0 \) and \( y = \frac{\sqrt{3}}{2} \). Thus, \( \sin^{-1}(0) - \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) = C \). Therefore, \( 0 - \frac{\pi}{3} = C \), giving \( C = -\frac{\pi}{3} \).

Step 4: Write the Implicit Solution

Substitute back the value of \( C \) into the solution: \( \sin^{-1} x - \sin^{-1} y = C = -\frac{\pi}{3} \). Solving, the implicit solution is: \( \sin^{-1} x - \sin^{-1} y = \pi/3 \).

Step 5: Solve for y in Terms of x

To find \( y \) in terms of \( x \), use the solution \( \sin^{-1} x - \sin^{-1} y = \pi/3 \), rearranged to \( \sin^{-1} y = \sin^{-1} x - \pi/3 \).

Step 6: Apply Trigonometric Identity

Using the addition formula, \( y = \sin(\sin^{-1} x + \frac{\pi}{3}) \), solve using the identity \( \sin(a+b) = \sin a \cos b + \cos a \sin b \).

Step 7: Final Expression for y

This becomes \( y = x \cos \frac{\pi}{3} + \sqrt{1-x^2} \sin \frac{\pi}{3} \). With values \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \), then \( y = \frac{x}{2} + \frac{\sqrt{3} \sqrt{1-x^2}}{2} \).

Key Concepts

Separation of Variables

Separation of variables is a common technique used to solve differential equations. It involves rearranging an equation so that each variable is on a different side of the equation. To apply this method successfully, the equation must be one where all of the terms involving one variable can be entirely separated from the terms involving the other variable. In the problem given, the differential equation is initially expressed as:\[\frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0\]This equation can be rearranged by moving all terms involving \(x\) to one side and all terms involving \(y\) to the other side:\[\frac{d x}{\sqrt{1-x^{2}}} = \frac{d y}{\sqrt{1-y^{2}}}\]Once separated, each side of the equation is integrated independently. This results in two expressions, one for each variable, with a constant of integration usually denoted as \(C\). In this case:\[\sin^{-1} x = \sin^{-1} y + C\] Remember, the integration constant \(C\) can be determined if initial conditions are provided.

Initial Value Problem

An initial value problem is a type of differential equation problem where you are given a differential equation along with specific values at the start, known as initial conditions. These conditions are used to find the particular solution that fits the given data. In the exercise, the initial conditions provided are:

• \(x = 0\)
• \(y = \frac{\sqrt{3}}{2}\)

These values allow us to find the constant \(C\) in the equation resulting from integrating after separating the variables.We plug these values into the equation:\[\sin^{-1}(0) - \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) = C\]Since \( \sin^{-1}(0) = 0 \) and \( \sin^{-1}\left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \), the value of \(C\) can be calculated:\[0 - \frac{\pi}{3} = C \C = -\frac{\pi}{3}\]Adding this result back to our equation gives the specific solution satisfying the initial condition.

Trigonometric Identities

Trigonometric identities are essential tools in calculus and differential equations. They help simplify complex expressions and solve equations involving trigonometric functions. In this particular exercise, we encounter the identity for sine of a sum:\[\sin(a + b) = \sin a \cos b + \cos a \sin b\]This identity is useful when solving the differential equation for \(y\) in terms of \(x\). After establishing the relationship:\[\sin^{-1} y = \sin^{-1} x - \frac{\pi}{3}\]We use the identity by writing:\[y = \sin(\sin^{-1} x + \frac{\pi}{3})\]Substituting \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\), we arrive at:\[y = x \cdot \frac{1}{2} + \sqrt{1-x^2} \cdot \frac{\sqrt{3}}{2}\y = \frac{x}{2} + \frac{\sqrt{3} \sqrt{1-x^2}}{2}\]This application of trigonometric identities simplifies the expression, making it easier to work with or evaluate further. Understanding these identities is essential for handling similar problems efficiently.