The standard form of a linear firs-order DE is d y </mrow> d

The standard form of a linear firs-order DE is
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
I think the equation is separable if and only if $P\left(x\right)$ and $Q\left(x\right)$ are constants, but I'm not sure. (Haven't found any counterexamples but also can't seem to prove it.) Can anyone confirm or deny that this is correct?
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bluayu0y
For $P=0$ and for any $Q$ The DE is separable. You have also the case when $P=\lambda Q$ Where $\lambda$ is a constant, then it's also separable:
${y}^{\prime }+P\left(x\right)y=Q\left(x\right)$
${y}^{\prime }+\lambda Q\left(x\right)y=Q\left(x\right)$
${y}^{\prime }=Q\left(x\right)\left(1-\lambda y\right)$
$\frac{dy}{1-\lambda y}=Q\left(x\right)dx$
Not exactly what you’re looking for?
hawwend8u
Your assertion is not true. For example, if Q(x)=0, regardless of P(x), the equation is separable.