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If $\phi(\mathrm{x})$ is a differential function, then the solution of the differential equation $\mathrm{dy}+\left\{\mathrm{y} \phi^{\prime}(\mathrm{x})-\phi(\mathrm{x})\right.$
$\left.\phi^{\prime}(x)\right\} d x=0,$ is
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$\left.\phi^{\prime}(x)\right\} d x=0,$ is
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Verified Answer
The correct answer is:
$y=\{\phi(x)-1\}+C e^{-\phi(x)}$
Given differential equation is
$\mathrm{dy}+\left\{y \phi^{\prime}(x)-\phi(x) \phi^{\prime}(x)\right\} d x=0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\phi^{\prime}(\mathrm{x}) \mathrm{y}=\phi(\mathrm{x}) \phi^{\prime}(\mathrm{x})$
which is a linear differential equation with $\quad P=\phi^{\prime}(x), Q=\phi(x) \cdot \phi^{\prime}(x)$ and
$I \cdot F=e^{\int \phi^{\prime}(x) d x}=e^{\phi(x)}$
$\therefore$ Solution is $y \cdot e^{\phi(x)}=\int \phi(x) \cdot \phi^{\prime}(x) e^{\phi(x)} d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\int \phi(x) \cdot e^{\phi(x)} \phi^{\prime}(x) d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-\int \phi^{\prime}(x) e^{\phi(x)} d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-e^{\phi(x)}+C$
$\Rightarrow y=\lceil\phi(x)-1]+C e^{-\phi(x)}$
$\mathrm{dy}+\left\{y \phi^{\prime}(x)-\phi(x) \phi^{\prime}(x)\right\} d x=0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\phi^{\prime}(\mathrm{x}) \mathrm{y}=\phi(\mathrm{x}) \phi^{\prime}(\mathrm{x})$
which is a linear differential equation with $\quad P=\phi^{\prime}(x), Q=\phi(x) \cdot \phi^{\prime}(x)$ and
$I \cdot F=e^{\int \phi^{\prime}(x) d x}=e^{\phi(x)}$
$\therefore$ Solution is $y \cdot e^{\phi(x)}=\int \phi(x) \cdot \phi^{\prime}(x) e^{\phi(x)} d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\int \phi(x) \cdot e^{\phi(x)} \phi^{\prime}(x) d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-\int \phi^{\prime}(x) e^{\phi(x)} d x+C$
$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-e^{\phi(x)}+C$
$\Rightarrow y=\lceil\phi(x)-1]+C e^{-\phi(x)}$
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