Let $e^{αy} + e^{βy} + γx^2 +δ\log |x| + C = 0$, where $C∈R$ 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 $(α+β+γ+δ-C)$ is

$e+\frac{1}{e}$

The correct answer is Option (3) → $e+\frac{1}{e}$

Given differential equation

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

Divide both sides by $e^y$

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

Separate variables

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

Divide by $x$

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

Integrate both sides

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

$e^y+e^{-y}=-\left(\frac{x^2}{2}-\log|x|\right)+C$

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

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

$\alpha=1,\;\beta=-1,\;\gamma=\frac12,\;\delta=-1$

Use point $(1,1)$

$e+e^{-1}+\frac12-\log1+C=0$

$C=-(e+e^{-1}+\frac12)$

Compute

$\alpha+\beta+\gamma+\delta-C=1-1+\frac12-1-[-(e+e^{-1}+\frac12)]$

$=-\frac12+e+e^{-1}+\frac12$

$=e+e^{-1}$

The value of $\alpha+\beta+\gamma+\delta-C$ is $e+\frac{1}{e}$.