If the slope of the tangent of the curve at any point is equal to $-y+\mathrm{e}^{-x}$, then the equation of…
If the slope of the tangent of the curve at any point is equal to $-y+\mathrm{e}^{-x}$, then the equation of the curve passing through origin is
- $y+x \mathrm{e}^x=0$
- $y \mathrm{e}^x+x=0$
- $y \mathrm{e}^x-x=0$
- $y-x \mathrm{e}^x=0$
Solution
$\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=-y+\mathrm{e}^{-x} \\
\Rightarrow & \frac{\mathrm{d} y}{\mathrm{~d} x}+y=\mathrm{e}^{-x} \\
\therefore \quad & \text { I.F. }=\mathrm{e}^{\mathrm{dx}}=\mathrm{e}^x
\end{aligned}$
$\therefore \quad$ Solution of the given equation is
$\begin{gathered}
y \mathrm{e}^x=\int \mathrm{e}^x \cdot \mathrm{e}^{-x} \mathrm{~d} x+\mathrm{c} \\
\Rightarrow y \mathrm{e}^x=\int \mathrm{d} x \pm \mathrm{c} \\
\Rightarrow y \mathrm{e}^x=x+\mathrm{c}
\end{gathered}$
Since the curve passes through $(0,0)$.
$\begin{array}{ll}
\therefore \quad & 0=0+\mathrm{c} \\
& \Rightarrow \mathrm{c}=0 \\
\therefore \quad & y \mathrm{e}^x=x \\
& \Rightarrow Y \mathrm{e}^x-x=0
\end{array}$
Asked in: MHT CET 2023 (14 May Shift 2)
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