Differential Equations Question 11
Question 11 - 01 February - Shift 2
Let $\alpha x=\exp (x^{\beta} y^{\gamma})$ be the solution of the differential equation $2 x^{2} y$ dy $-(1-x y^{2}) d x=0$, $x>0, y(2)=\sqrt{\log _{e} 2}$. Then $\alpha+\beta-\gamma$ equals :
(1) 1
(2) -1
(3) 0
(4) 3
Show Answer
Answer: (1)
Formula: Equations reducible to linear form (BERNOULI’S EQUATION)
$ \begin{aligned} & \alpha x=\mathrm{e}^{\mathrm{x}^\beta \cdot \mathrm{y}^\gamma} \\ & 2 \mathrm{x}^2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1-\mathrm{x} \cdot \mathrm{y}^2 \quad \mathrm{y}^2=\mathrm{t} \\ & \mathrm{x}^2 \frac{\mathrm{dt}}{\mathrm{dx}}=1-\mathrm{xt} \\ & \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{\mathrm{x}}=\frac{1}{\mathrm{x}^2} \\ & \text { I.F. }=\mathrm{e}^{\ln \mathrm{x}}=\mathrm{x} \\ & \mathrm{t}(\mathrm{x})=\int \frac{1}{\mathrm{x}^2} \cdot \mathrm{x} d \mathrm{x} \\ & \mathrm{y}^2 \cdot \mathrm{x}=\ln \mathrm{x}+\mathrm{C} \\ & \therefore 2 . \ell \mathrm{n} 2=\ell \mathrm{n} 2+\mathrm{C} \\ & \therefore \mathrm{C}=\ell \mathrm{n} 2 \\ & \end{aligned} $
Hence, $\mathrm{xy}^2=\ell \mathrm{n} 2 \mathrm{x}$ $ \therefore 2 \mathrm{x}=\mathrm{e}^{\mathrm{x} \cdot \mathrm{y}^2} $
Hence $\alpha=2, \beta=1, \gamma=2$
