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
Answer & explanation
Correct answer: option 3
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 $e+\frac{1}{e}$.