JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 7)
If the solution curve $$f(x, y)=0$$ of the differential equation
$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,
passes through the points $$(1,0)$$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :
$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,
passes through the points $$(1,0)$$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :
$$e^{\sqrt{2} e^{2}}$$
$$e^{2 e^{\sqrt{2}}}$$
$$e^{e^{2}}$$
$$e^{2 e^{2}}$$
Explanation
We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$
Put $x \log _e x=t$
$$ \begin{aligned} & \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \therefore \frac{d t}{d y}-t =e^{y} \end{aligned} $$
Now, IF $=e^{\int-d y}=e^{-y}$
$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$
Equation (i), passes through the point $(1,0)$
$$ \begin{aligned} & \therefore 0=0+c \\\\ & \Rightarrow c=0 \\\\ & \therefore \left(x \log _e x\right) e^{-y}=y \\\\ & \Rightarrow x \log _e x=y e^y \end{aligned} $$
Which passes through $(\alpha, 2)$
$$ \begin{aligned} & \therefore \alpha \log _e \alpha=2 e^2 \\\\ & \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\ & \Rightarrow \alpha^\alpha=e^{2 e^2} \end{aligned} $$
Put $x \log _e x=t$
$$ \begin{aligned} & \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\ & \therefore \frac{d t}{d y}-t =e^{y} \end{aligned} $$
Now, IF $=e^{\int-d y}=e^{-y}$
$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$
Equation (i), passes through the point $(1,0)$
$$ \begin{aligned} & \therefore 0=0+c \\\\ & \Rightarrow c=0 \\\\ & \therefore \left(x \log _e x\right) e^{-y}=y \\\\ & \Rightarrow x \log _e x=y e^y \end{aligned} $$
Which passes through $(\alpha, 2)$
$$ \begin{aligned} & \therefore \alpha \log _e \alpha=2 e^2 \\\\ & \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\ & \Rightarrow \alpha^\alpha=e^{2 e^2} \end{aligned} $$
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