JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 11)
If x = x(y) is the solution of the differential equation
$$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$$; then x(e) is equal to :
$$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$$; then x(e) is equal to :
$${e^3}({e^e} - 1)$$
$${e^e}({e^3} - 1)$$
$${e^2}({e^e} + 1)$$
$${e^e}({e^2} - 1)$$
Explanation
$\frac{d x}{d y}-\frac{2 x}{y}=y^{2}(y+1) e^{y}$
$$ \text { If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}} $$
Solution is given by
$$ \begin{aligned} &x \cdot \frac{1}{y^{2}}=\int y^{2}(y+1) e^{y} \cdot \frac{1}{y^{2}} d y \\\\ \Rightarrow & \frac{x}{y^{2}}=\int(y+1) e^{y} d y \\\\ \Rightarrow & \frac{x}{y^{2}}=y e^{y}+c \end{aligned} $$
$\Rightarrow x=y^{2}\left(y e^{y}+c\right)$ at, $y=1, x=0$
$\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e$ at $y=e$,
$x=e^{2}\left(e . e^{e}-e\right)$
$$ \text { If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}} $$
Solution is given by
$$ \begin{aligned} &x \cdot \frac{1}{y^{2}}=\int y^{2}(y+1) e^{y} \cdot \frac{1}{y^{2}} d y \\\\ \Rightarrow & \frac{x}{y^{2}}=\int(y+1) e^{y} d y \\\\ \Rightarrow & \frac{x}{y^{2}}=y e^{y}+c \end{aligned} $$
$\Rightarrow x=y^{2}\left(y e^{y}+c\right)$ at, $y=1, x=0$
$\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e$ at $y=e$,
$x=e^{2}\left(e . e^{e}-e\right)$
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