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

\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=-y+\mathrm{e}^{-x} \\
\Rightarrow & \frac{\mathrm{d} y}{\mathrm{~d} x}+y=\mathrm{e}^{-x} \\
\therefore \quad & \text { I.F. }=\mathrm{e}^{\mathrm{dx}}=\mathrm{e}^x
\end{aligned}\)
\(\therefore \quad\) 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 \((0,0)\).
\(\begin{array}{ll}
\therefore \quad & 0=0+\mathrm{c} \\
& \Rightarrow \mathrm{c}=0 \\
\therefore \quad & y \mathrm{e}^x=x \\
& \Rightarrow y \mathrm{e}^x-x^2=0
\end{array}\)