JEE Main 2019 (Online) 12th April Morning ShiftQuestion 1
Consider the differential equation, $y^2dx+\left(x-\frac1y\right)dy=0$. If value of $y$ is $1$ when $x=1$, then the value of $x$ for which $y=2$, is
Differential Equation
Solution
Concept. Divide by $y^2\,dy$: linear in $x$ with IF $e^{-1/y}$. Formulas used. (1) $\int\dfrac{e^{-1/y}}{y^3}\,dy=\left(1+\dfrac1y\right)e^{-1/y}$ (put $u=-\frac1y$)Step 1 - Linear form. $\dfrac{dx}{dy}+\dfrac{x}{y^2}=\dfrac{1}{y^3}$, IF $=e^{\int y^{-2}dy}=e^{-1/y}$. Step 2 - Integrate. By (1), $x\,e^{-1/y}=\left(1+\frac1y\right)e^{-1/y}+C$, so $$x=1+\frac1y+C\,e^{1/y}.$$ Step 3 - Use $x=1$ at $y=1$. $1=2+Ce$, so $C=-\frac1e$. Step 4 - At $y=2$. $x=1+\frac12-\frac{e^{1/2}}{e}=\frac32-\frac{1}{\sqrt e}$. Answer: (A)