JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
If $x=x(y)$ is the solution of the differential equation $y\frac{dx}{dy}=2x+y^{3}(y+1)e^{y}$, $x(1)=0$; then $x(e)$ is equal to :
Differential Equation
Solution
Concept. Linear in $x$; IF $\frac{1}{y^2}$. Formulas used. (1) $\left(\dfrac{x}{y^2}\right)'=\dfrac{1}{y^2}\left(x'-\dfrac2yx\right)$(2) $\int(y+1)e^y\,dy=ye^y$Step 1 - Linear form. $x'-\dfrac2y\,x=y^2(y+1)e^y$. Step 2 - Solve. By (1) and (2), $\dfrac{x}{y^2}=ye^y+C$. Step 3 - Constant. $x(1)=0$: $0=e+C$, so $C=-e$. Step 4 - At $y=e$. $x(e)=e^2\left(e\cdot e^e-e\right)=e^3\left(e^e-1\right)$. Answer: (A)