Discussion
Bernoulli's equation \( \dfrac{dy}{dx} + Py = Qy^{n} \) is made linear by the substitution
Answer: A
Dividing through by \( y^{n} \) leaves \( y^{-n}y' + Py^{1-n} = Q \), and with \( v = y^{1-n} \) we get \( v' = (1-n)y^{-n}y' \), so
\[ \frac{1}{1-n}\frac{dv}{dx} + Pv \;=\; Q, \]
which is linear in \( v \). The substitution \( v = \ln y \) suits an equation with \( n = 1 \), which is separable anyway
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