JEE Main 2023 (Online) 1st February Evening ShiftQuestion 1
Let $\alpha x=\exp\left(x^{\beta}y^{\gamma}\right)$ be the solution of the differential equation $2x^{2}y\,dy-\left(1-xy^{2}\right)dx=0$, $x>0$, $y(2)=\sqrt{\log_e2}$. Then $\alpha+\beta-\gamma$ equals :
Differential Equation
Solution
Concept. $2yy'=(y^2)'$; with $w=y^2$ the left side becomes $x(xw)'$. Formulas used. (1) $w=y^2\Rightarrow w'=2yy'$(2) $(xw)'=xw'+w$Step 1 - Substitute. By (1), $x^2w'=1-xw$, i.e. $x^2w'+xw=1$. Step 2 - Integrate. By (2), $x(xw)'=1$, so $(xw)'=\frac1x$ and $xy^2=\ln x+C$. Step 3 - Constant. $y(2)^2=\ln2$: $2\ln2=\ln2+C$, so $C=\ln2$ and $xy^2=\ln(2x)$. Step 4 - Compare. $2x=e^{xy^2}$, so $\alpha=2$, $\beta=1$, $\gamma=2$ and $\alpha+\beta-\gamma=1$. Answer: (A)