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=e^{x^\beta \cdot y^\gamma} \\ & 2 x^2 y \frac{dy}{dx}=1-x \cdot y^2 \quad y^2=t \\ & x^2 \frac{dt}{dx}=1-xt \\ & \frac{dt}{dx}+\frac{t}{x}=\frac{1}{x^2} \\ & \text { I.F. }=e^{\ln x}=x \\ & t(x)=\int \frac{1}{x^2} \cdot x d x \\ & y^2 \cdot x=\ln x+C \\ & \therefore 2 . \ell n 2=\ell n 2+C \\ & \therefore C=\ell n 2 \\ & \end{aligned} $
Hence, $xy^2=\ell n 2 x$ $ \therefore 2 x=e^{x \cdot y^2} $
Hence $\alpha=2, \beta=1, \gamma=2$
