JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
If $\frac{dy}{dx}=\frac{2^{x}y+2^{y}\cdot2^{x}}{2^{x}+2^{x+y}\log_e2}$, $y(0)=0$, then for $y=1$, the value of $x$ lies in the interval :
Differential Equation
Solution
Concept. After cancelling $2^x$, $\frac{dx}{dy}$ is (derivative of $y+2^y$)$/(y+2^y)$. Formulas used. (1) $\dfrac{d}{dy}\left(2^y\right)=2^y\ln2$(2) $\int\dfrac{f'(y)}{f(y)}\,dy=\ln|f(y)|+C$Step 1 - Simplify. $$\frac{dy}{dx}=\frac{2^x\left(y+2^y\right)}{2^x\left(1+2^y\ln2\right)}=\frac{y+2^y}{1+2^y\ln2}.$$ Step 2 - Flip and integrate. By (1), the numerator of $\frac{dx}{dy}=\frac{1+2^y\ln2}{y+2^y}$ is the derivative of the denominator. By (2), $$x=\ln\left(y+2^y\right)+C.$$ Step 3 - Use $y(0)=0$. $0=\ln1+C$, so $C=0$. Step 4 - At $y=1$. $x=\ln3\approx1.10$, which lies in $(1,2)$. Answer: (C)