The solution of $\frac{d y}{d x}+P(x) y=0$, is

$y=C e^{\int P d x}$ $y=C e^{-\int P d x}$ $x=C e^{-\int P d y}$ $x=C e^{\int P d y}$

$y=C e^{-\int P d x}$

We have, $\frac{d y}{d x}+P(x) y=0 \Rightarrow d y+P(x) y d x=0 \Rightarrow \frac{d y}{y}=-P(x) d x$ On integrating, we get $\log y=\int-P(x) d x+\log C \Rightarrow y=C e^{\int-P d x}$