Objective: Solve a first-order differential equation. This involves finding the general solution using separation of variables, determining the particular solution using a given point (1, 1), and then comparing coefficients with a specified form to find \(\alpha, \beta, \gamma, \delta\), and C. Finally, compute the expression \((\alpha + \beta + \gamma + \delta - C)\).
Differential Equation:\[ x(e^{2y} - 1)dy + (x^2 - 1)e^y dx = 0 \]Method: Separation of variables. Rearrange, group terms, integrate, and apply the initial condition.Solution Process:
1. Rearrange the equation: \(x(e^{2y} - 1)dy = -(x^2 - 1)e^y dx\)
2. Separate variables: \(\frac{e^{2y} - 1}{e^y} dy = -\frac{x^2 - 1}{x} dx\)
3. Simplify: \((e^y - e^{-y}) dy = -(x - \frac{1}{x}) dx\)
4. Integrate both sides: \(\int (e^y - e^{-y}) dy = -\int (x - \frac{1}{x}) dx\)
5. Perform integration: \(e^y + e^{-y} = -\left(\frac{x^2}{2} - \log|x|\right) + K\), where K is the constant of integration.
6. Rewrite the general solution: \(e^y + e^{-y} + \frac{1}{2}x^2 - \log|x| - K = 0\)
7. Compare with the given form \(e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0\) to find: \(\alpha = 1\), \(\beta = -1\), \(\gamma = \frac{1}{2}\), \(\delta = -1\), \(C = -K\).
8. Use the point (1, 1) to find K: \(e^1 + e^{-1} + \frac{1}{2}(1)^2 - \log|1| = K \implies K = e + \frac{1}{e} + \frac{1}{2}\).
9. Determine C: \(C = -K = -\left(e + \frac{1}{e} + \frac{1}{2}\right)\).
10. Compute the expression \((\alpha + \beta + \gamma + \delta - C)\): \(\left(1 + (-1) + \frac{1}{2} + (-1)\right) - \left(-\left(e + \frac{1}{e} + \frac{1}{2}\right)\right) = -\frac{1}{2} + e + \frac{1}{e} + \frac{1}{2} = e + \frac{1}{e}\).
Result: The value of \((\alpha + \beta + \gamma + \delta - C)\) is \(e + \frac{1}{e}\).