Question

Let \(e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0\), where \(C \in \mathbb{R}\) be a particular solution of the differential equation \(x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0\) and passes through the point (1, 1). The value of \((\alpha + \beta + \gamma + \delta - C)\) is

• A. \(e - 1\)
• B. \(e^2 - 1\)
• C. \(e + \frac{1}{e}\)
• D. \(\frac{1}{e}\)

Hint:

For variable separable differential equations, the main goal is to isolate all 'y' terms with 'dy' and all 'x' terms with 'dx' on opposite sides of the equation. Be careful with algebraic manipulations, especially with negative signs and logarithms.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves solving a first-order differential equation using variable separation. We will first find the general solution, then use the point (1, 1) to determine the particular solution. Finally, by comparing our solution's coefficients with the provided form \(e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0\), we will identify \(\alpha, \beta, \gamma, \delta\), and C, and then compute the required expression.
Step 2: Key Formula or Approach:
The differential equation to be solved is:\[ x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0 \]The method of separation of variables will be applied, followed by integration of both sides.
Step 3: Detailed Explanation:
Rearrange the differential equation:\[ x(e^{2y} - 1)dy = -(x^2 - 1)e^y dx \]Separate the variables by dividing by \(xe^y\):\[ \frac{e^{2y} - 1}{e^y} dy = -\frac{x^2 - 1}{x} dx \]Simplify the terms on both sides:\[ (e^y - e^{-y}) dy = -(x - \frac{1}{x}) dx \]Integrate both sides:\[ \int (e^y - e^{-y}) dy = -\int (x - \frac{1}{x}) dx \]This yields:\[ e^y + e^{-y} = -\left(\frac{x^2}{2} - \log|x|\right) + K \]where K is the constant of integration. Rearranging to match the target form:\[ e^y + e^{-y} + \frac{1}{2}x^2 - \log|x| - K = 0 \]Comparing this to \(e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0\), we find:\(\alpha = 1\)
\(\beta = -1\)
\(\gamma = \frac{1}{2}\)
\(\delta = -1\)
\(C = -K\)
Using the point (1, 1) to find K:\[ e^1 + e^{-1} + \frac{1}{2}(1)^2 - \log|1| = K \]\[ e + \frac{1}{e} + \frac{1}{2} = K \]Therefore, \(C = -K = -\left(e + \frac{1}{e} + \frac{1}{2}\right)\).Now, we compute the expression \((\alpha + \beta + \gamma + \delta - C)\):\[ = \left(1 + (-1) + \frac{1}{2} + (-1)\right) - \left(-\left(e + \frac{1}{e} + \frac{1}{2}\right)\right) \]\[ = \left(\frac{1}{2} - 1\right) + \left(e + \frac{1}{e} + \frac{1}{2}\right) \]\[ = -\frac{1}{2} + e + \frac{1}{e} + \frac{1}{2} \]\[ = e + \frac{1}{e} \]Step 4: Final Answer:
The calculated value of \((\alpha + \beta + \gamma + \delta - C)\) is \(e + \frac{1}{e}\).