Problem 18 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(e^{x}=y+\sqrt{1+y^{2}}\), then \(y\) is equal to (a) \(e^{x}+e^{-x}\) (b) \(e^{x}-e^{-x}\) (c) \(\frac{1}{2}\left(e^{x}-e^{-x}\right)\) (d) \(\frac{1}{2}\left(e^{x}+e^{-x}\right)\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\frac{1}{2}(e^x - e^{-x})\).

Step 1: Substitution and Squaring

We start with the equation \(e^{x}=y+ ext{√}(1+y^{2})\). To eliminate the square root, isolate it: \(y=e^x - ext{√}(1+y^{2})\). Next, square both sides of \(e^x = y + \text{√}(1+y^{2})\): \[(e^{x})^2 = (y + ext{√}(1+y^{2}))^2\].Expand the right side to get: \[e^{2x} = y^2 + 2y ext{√}(1+y^{2}) + (1+y^{2})\].

Step 2: Simplifying the Squared Equation

Upon simplifying the equation \[e^{2x} = y^2 + 2y ext{√}(1+y^{2}) + 1 + y^2\] , we get: \[e^{2x} = 2y^2 + 1 + 2y ext{√}(1+y^{2})\]. By rearranging terms and equating, the square root term can be isolated and squaring both sides again eliminates it, leading to a more solvable equation for \(y\).

Step 3: Isolating Terms for Simplification

Rearrange the expression \(e^{2x} = 1 + 2y^2 + 2y\sqrt{1+y^2}\) to isolate the square root term: \(e^{2x} - 2y^2 - 1 = 2y\text{√}(1+y^{2})\). Divide both sides by 2 to get:\[\frac{e^{2x} - 1}{2} = y\text{√}(1+y^{2})\].

Step 4: Solving for y

Using \(y=e^x-\text{√}(1+y^{2})\) and substituting \(y\) into the expression \(\frac{e^{2x} - 1}{2} = y\text{√}(1+y^{2})\), we find a potential \(y\) that fits with hyperbolic identities. Recognizing hyperbolic functions, the relation \(e^x - e^{-x} = 2y\) hints \(y = \frac{1}{2}(e^x - e^{-x})\).

Step 5: Verifying the Solution

Upon noticing that the setup resembles hyperbolic functions, we check whether \(y = \frac{1}{2}(e^x - e^{-x})\) satisfies the given original equation \(e^x = y + \sqrt{1+y^2}\). Simplifying both sides with this \(y\) value confirms the identity matches. Thus, the solution is verified.

Key Concepts

Differential Equations

Differential equations play a crucial role in modeling real-world phenomena. They are equations that involve derivatives of a function or functions. In essence, they describe how a quantity changes with respect to another variable, often time. This makes them indispensable in physics, engineering, economics, and biology. Differential equations can be categorized into various types:

• Ordinary Differential Equations (ODEs), where derivatives are taken with respect to a single variable.
• Partial Differential Equations (PDEs), which involve derivatives with respect to more than one variable.

In our current problem, we're dealing with implicit relations rather than a direct differential equation, but understanding the foundational concept of change they provide helps in solving such equations. This connection becomes clear as we apply techniques like square rooting and substitutions to simplify equations, much like solving differential forms.

Exponential Functions

Exponential functions are mathematical expressions in the form of \(f(x) = a \, e^{bx}\), where \(e\) is Euler's number, approximately 2.71828. These functions exhibit rapid growth or decay and are pivotal in mathematical modeling. Some key properties of exponential functions include:

• The base \(e\) means that the function's growth rate doubles for every unit increase in \(x\).
• They provide a smooth, continuous curve, essential for solving differential equations effectively.

In the problem given, the exponential function \(e^x\) is combined with other terms. This setup reflects hyperbolic function relationships. By recognizing connections of these functions through equations, such as \(e^x - e^{-x}\), the exact unknown quantity, like \(y\), can be simplified and identified. This ability to transform and manipulate exponential forms is vital in calculus and related fields.

Square Root Elimination

Square roots present unique challenges in equations, often complicating the path to a solution. The process of eliminating square roots involves careful manipulation and is a frequent step in solving mathematical problems. Here's a simplified outline of how square roots might be eliminated:

• Isolate the square root term in the equation whenever possible. This helps in focusing the operations you perform next.
• Square both sides of the equation to eliminate the square root. This step must be done carefully to avoid introducing extraneous solutions.

In the original problem, this method is utilized to simplify \(e^x = y + \sqrt{1+y^2}\). By squaring the equation, the square root is effectively eliminated. This simplification is critical for ease in solving and checking solutions, as seen when identifying the value of \(y\). Understanding such techniques is crucial for effectively manipulating algebraic expressions and solving complex equations.