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The general solution of the differential equation \elln $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\mathrm{x}=0$ is? $\quad$
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Verified Answer
The correct answer is:
$\mathrm{y}=-\mathrm{e}^{-\mathrm{x}}+\mathrm{c}$
Let $\ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\mathrm{x}=0 \Rightarrow \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=-\mathrm{x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}$
Integrate both the side, $y=-e^{-x}+c$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}$
Integrate both the side, $y=-e^{-x}+c$
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