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 $x(e^{2y} - 1)dy + (x^2 - 1)e^ydx = 0$ and passes through the point $(1, 1)$. The value of $(\alpha + \beta + \gamma + \delta - C)$ is

We are given:

$x(e^{2y} - 1)\,dy + (x^2 - 1)e^y\,dx = 0$

Our goal is to separate the variables. Notice that dividing the entire equation by $x \cdot e^y$ cleanly separates the $x$ and $y$ terms:

$\dfrac{e^{2y} - 1}{e^y}\,dy + \dfrac{x^2 - 1}{x}\,dx = 0$

Simplifying each fraction:

$\dfrac{e^{2y} - 1}{e^y} = \dfrac{e^{2y}}{e^y} - \dfrac{1}{e^y} = e^y - e^{-y}$

$\dfrac{x^2 - 1}{x} = x - \dfrac{1}{x}$

So we get:

$(e^y - e^{-y})\,dy + \left(x - \dfrac{1}{x}\right)dx = 0$

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$\int(e^y - e^{-y})\,dy + \int\left(x - \dfrac{1}{x}\right)dx = 0$

Integrating term by term:

$\int e^y\,dy = e^y$

$\int -e^{-y}\,dy = -(-e^{-y}) = +e^{-y}$ — don't miss this sign flip, since $\int e^{-y}\,dy = -e^{-y}$

$\int x\,dx = \dfrac{x^2}{2}$

$\int -\dfrac{1}{x}\,dx = -\log|x|$

Combining everything with a single constant of integration $C$:

$e^y + e^{-y} + \dfrac{x^2}{2} - \log|x| + C = 0$

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Comparing with $e^{\alpha y} + e^{\beta y} + \gamma x^2 + \delta\log|x| + C = 0$:

$\alpha = 1,\quad \beta = -1,\quad \gamma = \dfrac{1}{2},\quad \delta = -1$

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The curve passes through $(1, 1)$, so substituting $x = 1$ and $y = 1$:

$e^1 + e^{-1} + \dfrac{1}{2} - \underbrace{\log|1|}_{=\,0} + C = 0$

$e + \dfrac{1}{e} + \dfrac{1}{2} + C = 0$

$C = -e - \dfrac{1}{e} - \dfrac{1}{2}$

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$\alpha + \beta + \gamma + \delta - C$

$= 1 + (-1) + \dfrac{1}{2} + (-1) - \left(-e - \dfrac{1}{e} - \dfrac{1}{2}\right)$

$= -\dfrac{1}{2} + e + \dfrac{1}{e} + \dfrac{1}{2}$

$= e + \dfrac{1}{e}$