Let \(e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta \log|x| + C = 0\), where \(C \in \mathrm{R}\), 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
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(e + \displaystyle\frac{1}{e}\)
The equation is in variables separable form. Dividing by \(x e^y\):
\( \displaystyle \frac{e^{2y} - 1}{e^y}\,dy + \frac{x^2 - 1}{x}\,dx = 0 \)
\( \displaystyle (e^y - e^{-y})\,dy + \left(x - \frac{1}{x}\right)dx = 0 \)
Integrating:
\( \displaystyle e^y + e^{-y} + \frac{x^2}{2} - \log|x| + C = 0 \)
Comparing with the given form (the two exponentials can be taken in either order; only \(\alpha + \beta\) is needed):
\( \displaystyle \alpha + \beta = 1 + (-1) = 0,\quad \gamma = \frac{1}{2},\quad \delta = -1 \)
The curve passes through (1, 1), and \(\log 1 = 0\):
\( \displaystyle e + \frac{1}{e} + \frac{1}{2} + C = 0 \)
\( \displaystyle C = -\left(e + \frac{1}{e} + \frac{1}{2}\right) \)
Hence:
\( \displaystyle \alpha + \beta + \gamma + \delta - C = 0 + \frac{1}{2} - 1 + e + \frac{1}{e} + \frac{1}{2} \)
\( \displaystyle = e + \frac{1}{e} \)