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By multiplying with $e^{\int P d x}$ on both sides of the equation $\frac{d y}{d x}+P(x) y=Q(x)$, the left side of the equation takes the form $\frac{d}{d x}(y f(x))$, then $f(x)=$
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The correct answer is:
$e^{\int P d x}$
$\frac{d y}{d x}+P(x) y=Q(x)$
multiplying $e^{\int P d x}$ on both sides of eqn.
$e^{\int P d x}$ $\frac{d y}{d x}+y$ $e^{\int P d x} P(x)$ $=Q(x) e^{\int P d x}$
equating LHS with $\frac{d}{d x}(y(f(x)))$
$\Rightarrow e^{\int P d x} \frac{d y}{d x}+$ $y e^{\int P d x} P(x)$ $=\frac{d}{d x}(y f(x))$
$\Rightarrow \quad \frac{d}{d x}\left(y e^{\int P d x}\right)$ $=\frac{d}{d x} y f(x)$
$f(x)=e^{\int P d x}$
multiplying $e^{\int P d x}$ on both sides of eqn.
$e^{\int P d x}$ $\frac{d y}{d x}+y$ $e^{\int P d x} P(x)$ $=Q(x) e^{\int P d x}$
equating LHS with $\frac{d}{d x}(y(f(x)))$
$\Rightarrow e^{\int P d x} \frac{d y}{d x}+$ $y e^{\int P d x} P(x)$ $=\frac{d}{d x}(y f(x))$
$\Rightarrow \quad \frac{d}{d x}\left(y e^{\int P d x}\right)$ $=\frac{d}{d x} y f(x)$
$f(x)=e^{\int P d x}$
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