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 requires solving a first-order differential equation using the method of separation of variables. After finding the general solution, we use the given point (1, 1) to find the particular solution. Finally, we compare the coefficients of our solution with the given form to determine the values of \(\alpha, \beta, \gamma, \delta\), and C, and compute the required expression.
Step 2: Key Formula or Approach:
The given differential equation is: \[ x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0 \] We will separate the variables (terms with y and dy on one side, terms with x and dx on the other) and then integrate both sides.
Step 3: Detailed Explanation:
Start by rearranging the equation: \[ x(e^{2y} - 1)dy = -(x^2 - 1)e^y dx \] Now, group the y-terms on the left and x-terms on the right. Divide by \(xe^y\): \[ \frac{e^{2y} - 1}{e^y} dy = -\frac{x^2 - 1}{x} dx \] Simplify both sides: \[ \left(\frac{e^{2y}}{e^y} - \frac{1}{e^y}\right) dy = -\left(\frac{x^2}{x} - \frac{1}{x}\right) dx \] \[ (e^y - e^{-y}) dy = -(x - \frac{1}{x}) dx \] Integrate both sides: \[ \int (e^y - e^{-y}) dy = -\int (x - \frac{1}{x}) dx \] \[ e^y - (-e^{-y}) = -\left(\frac{x^2}{2} - \log|x|\right) + K \] where K is the constant of integration. \[ e^y + e^{-y} = -\frac{x^2}{2} + \log|x| + K \] To match the given form, let's move all terms to one side: \[ e^y + e^{-y} + \frac{1}{2}x^2 - \log|x| - K = 0 \] Now, compare this with the given form \(e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0\). By comparing the coefficients, we get: \(\alpha = 1\)
\(\beta = -1\)
\(\gamma = \frac{1}{2}\)
\(\delta = -1\)
\(C = -K\)
The solution passes through the point (1, 1). Substitute x=1, y=1 into our solution to find K: \[ e^1 + e^{-1} + \frac{1}{2}(1)^2 - \log|1| = K \] \[ e + \frac{1}{e} + \frac{1}{2} - 0 = K \] So, \(K = e + \frac{1}{e} + \frac{1}{2}\). The constant C in the given form is \(C = -K\), so: \[ C = -\left(e + \frac{1}{e} + \frac{1}{2}\right) \] We need to find the value of \((\alpha + \beta + \gamma + \delta - C)\): \[ = \left(1 + (-1) + \frac{1}{2} + (-1)\right) - \left(-\left(e + \frac{1}{e} + \frac{1}{2}\right)\right) \] \[ = \left(0 + \frac{1}{2} - 1\right) + \left(e + \frac{1}{e} + \frac{1}{2}\right) \] \[ = -\frac{1}{2} + e + \frac{1}{e} + \frac{1}{2} \] \[ = e + \frac{1}{e} \] Step 4: Final Answer:
The value of the expression \((\alpha + \beta + \gamma + \delta - C)\) is \(e + \frac{1}{e}\).