Problem 28 - Advanced Engineering Mathematics

Question

From \(\frac{1}{1+(2 y)^{2}} d y=\frac{-x}{1+\left(x^{2}\right)^{2}} d x\) we obtain \(\frac{1}{2} \tan ^{-1} 2 y=-\frac{1}{2} \tan ^{-1} x^{2}+c \quad\) or \(\quad \tan ^{-1} 2 y+\tan ^{-1} x^{2}=c_{1}\). Using \(y(1)=0\) we find \(c_{1}=\pi / 4\). Thus, an implicit solution of the initial-value problem is \(\tan ^{-1} 2 y+\tan ^{-1} x^{2}=\pi / 4 .\) Solving for \(y\) and using a trigonometric identity we get $$\begin{aligned} 2 y &=\tan \left(\frac{\pi}{4}-\tan ^{-1} x^{2}\right) \\ y &=\frac{1}{2} \tan \left(\frac{\pi}{4}-\tan ^{-1} x^{2}\right) \\ &=\frac{1}{2} \frac{\tan \frac{\pi}{4}-\tan \left(\tan ^{-1} x^{2}\right)}{1+\tan \frac{\pi}{4} \tan \left(\tan ^{-1} x^{2}\right)} \\ &=\frac{1}{2} \frac{1-x^{2}}{1+x^{2}}. \end{aligned}$$

Step-by-Step Solution

Answer

The implicit solution is \( y = \frac{1}{2} \frac{1-x^2}{1+x^2} \).

Step 1: Rewrite the Differential Equation

We start with the equation \( \frac{1}{1+(2y)^{2}} dy = \frac{-x}{1+(x^{2})^{2}} dx \). This is a separable differential equation that can be solved by integrating both sides.

Step 2: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \). The integrals are as follows:\[ \text{Left: } \int \frac{1}{1+(2y)^{2}} dy = \frac{1}{2} \tan^{-1}(2y) + C_1 \]\[ \text{Right: } \int \frac{-x}{1+(x^{2})^{2}} dx = -\frac{1}{2} \tan^{-1}(x^2) + C_2 \]

Step 3: Equate Integrals

Equating the two integrations, we have:\[ \frac{1}{2} \tan^{-1}(2y) = -\frac{1}{2} \tan^{-1}(x^2) + C \]

Step 4: Simplify to Find General Solution

Multiply through by 2 to clear the fractions:\[ \tan^{-1}(2y) + \tan^{-1}(x^2) = C_1 \]

Step 5: Use Initial Condition

Given the initial condition \( y(1) = 0 \), substitute \( x = 1 \) and \( y = 0 \) into the equation:\[ \tan^{-1}(0) + \tan^{-1}(1^2) = C_1 \]\( \tan^{-1}(0) = 0 \) and \( \tan^{-1}(1) = \frac{\pi}{4} \), hence:\[ C_1 = \frac{\pi}{4} \]

Step 6: Implicit Solution

Substitute \( C_1 = \frac{\pi}{4} \) into the general solution to get:\[ \tan^{-1}(2y) + \tan^{-1}(x^2) = \frac{\pi}{4} \]

Step 7: Solve for \( y \) Using Trigonometric Identity

Solve for \( 2y \) using the identity \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \):\[ 2y = \tan \left( \frac{\pi}{4} - \tan^{-1}(x^2) \right) \]

Step 8: Use Tangent Identity to Simplify

Using the identity for \( \tan(A - B) \), solve for \( y \):\[ y = \frac{1}{2} \frac{\tan(\frac{\pi}{4}) - \tan(\tan^{-1}(x^2))}{1 + \tan(\frac{\pi}{4})\tan(\tan^{-1}(x^2))} \]Recognizing \( \tan(\frac{\pi}{4}) = 1 \), and \( \tan(\tan^{-1}(x^2)) = x^2 \), we simplify to:\[ y = \frac{1}{2} \frac{1 - x^2}{1 + x^2} \]

Step 9: Conclude Solution

The implicit solution, rearranged into an explicit form for \( y \), is:\[ y = \frac{1}{2} \frac{1 - x^2}{1 + x^2} \]

Key Concepts

Trigonometric Identities in Differential Equations

Trigonometric identities play a crucial role in solving differential equations, especially when simplifying expressions. In this problem, we use the trigonometric identity for the tangent of the difference of two angles:

• \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

This identity helps us express the implicit solution of the differential equation in a simpler form.
For example, in our solution, \( 2y = \tan\left(\frac{\pi}{4} - \tan^{-1}(x^2)\right) \) was simplified using this identity.
It's essential to recognize that the inverse tangent function \( \tan^{-1} \) is involved here.
So when you encounter such equations, look for appropriate trigonometric identities to help you manage complex expressions and derive cleaner results.
Remember, learning these identities can significantly boost your problem-solving skills in calculus and beyond.

Understanding Implicit Solutions

In solving differential equations, particularly those involving functions that are not easily isolated, an implicit solution can be very useful. An implicit solution is an equation where the dependent and independent variables are not separated.

• In this problem, the expression \( \tan^{-1}(2y) + \tan^{-1}(x^2) = \frac{\pi}{4} \) is an implicit solution.

This means we have an equation that defines a relationship between \( y \) and \( x \), but \( y \) is not explicitly solved out in terms of \( x \).
Why use implicit solutions? They can be more general and easier to find, especially when an explicit solution is too complex or impossible to capture with elementary functions.
While explicit solutions are easier to work with, mastering implicit solutions allows you to handle a broader class of problems, making them an invaluable tool in differential equations.

Initial Value Problem Explained

An initial value problem (IVP) in the context of differential equations requires solving the equation with specific conditions that the solution must satisfy. These conditions, known as initial conditions, specify the value of the solution at a particular point.

• In our example, you have the initial condition \( y(1) = 0 \). This is where the solution passes through the point when \( x = 1 \).

Initial conditions are crucial because they determine the constant \( C \) that appears after integration.
They effectively "lock in" a particular "leaf" of solutions from the infinite family of possible solutions given by the differential equation.
By solving for \( C \), you can find a specific solution that meets the required conditions.
This approach is important in many practical applications because it ensures the solution is aligned with real-world constraints or measurements.
As seen in our example, substituting \( y(1) = 0 \) into the simplified implicit solution helps us determine the constant \( C_1 \) as \( \frac{\pi}{4} \), thus refining our solution to the problem's specific needs.