Question

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

• A. 0
• B. -1
• C. 1
• D. 3

Answer 1

Answer: (C)

$\alpha x=e^{x^{\beta }\cdot y^{\gamma }}$
$2x^{2}y\frac{dy}{dx}=1-x\cdot y^2{y}^2=t$
$x^2\frac{dt}{dx}=1-xt$
$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{x^2}\text{ I.F. }=e^{\ell nx}=x$
$t(x)=\int \frac{1}{x^2}\cdot xdx$
$y^2\cdot x=\ell nx+C$
$\therefore 2\cdot \ell n2=\ell n2+C$
$\therefore C=\ell n2$
Hence, $x{y}^2=\ell n2x$
$\therefore 2x=e^{x\cdot y^2}$
Hence $\alpha =2,\beta =1,\gamma =2$