Question

Let $\alpha x=\exp \left(x^\beta y^{\gamma}\right)$ be the solution of the differential equation $2 x^2 y d y-\left(1-x y^2\right) d x=0, x >0$, $y(2)=\sqrt{\log _e 2}$. Then $\alpha+\beta-\gamma$ equals :

• A. 0
• B. -1
• C. 1
• D. 3

Answer 1

Answer: (C)

$ \alpha x=e^{x^\beta \cdot y^\gamma} $
$ 2 x^2 y \frac{d y}{d x}=1-x \cdot y^2 \,\,\,\, y^2=t $
$ x^2 \frac{d t}{d x}=1-x t $
$ \frac{d t}{d x}+\frac{t}{x}=\frac{1}{x^2} \,\,\,\, \text { I.F. }=e^{\ell n x}=x $
$ t(x)=\int \frac{1}{x^2} \cdot x d x $
$ y^2 \cdot x=\ell n x+C $
$ \therefore 2 \cdot \ell n 2=\ell n 2+C$
$\therefore C =\ell n 2$
Hence, $xy ^2=\ell n 2 x$
$\therefore 2 x = e ^{ x \cdot y ^2}$
Hence $\alpha=2, \beta=1, \gamma=2$