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The integrating factor of the linear differential equation $\frac{d y}{d x}=\frac{1}{4 x+3 y}$ is
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The correct answer is:
$\mathrm{e}^{-4 \mathrm{y}}$
Given $\frac{d y}{d x}=\frac{1}{4 x+3 y}$
$\frac{d x}{d y}=3 y+4 x \Rightarrow \frac{d x}{d y}-4 x=3 y$
Then, IF $=e^{\int-4 d y}=e^{-4 y}$
$\frac{d x}{d y}=3 y+4 x \Rightarrow \frac{d x}{d y}-4 x=3 y$
Then, IF $=e^{\int-4 d y}=e^{-4 y}$
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