# Bernoulli differential equation proving

As we know, the differential equation in the form is called the Bernoulli equation

$\frac {dy}{dx} + p(x)y = q(x)y^n$

How do i show that if $y$ is the solution of the above Bernoulli equation and $u = y^{1-n}$, then u satisfies the linear differential equation

$\frac{du}{dx} +(1-n)p(x)u = (1-n)q(x)$

I can use the substituion to use solve differential equations like

$y' + xy = xy^2$

but have no idea how to prove this question .

• Express $y$ in terms of $u$ and subsitute in the initial equation. – Yves Daoust Nov 1 '16 at 8:52
Hint. One may observe that from $u=y^{1-n}$, by the chain rule, we get $$\frac{du}{dx}=(1-n)\cdot\frac{dy}{dx}\cdot y^{-n}\qquad \text{or} \qquad \frac{dy}{dx}=\frac1{(1-n)}\cdot y^{n}\cdot\frac{du}{dx}$$ then plugging it into $$\frac {dy}{dx} + p(x)y = q(x)y^n$$ using $y=y^n u$ gives $$\frac1{(1-n)}\cdot y^{n}\cdot\frac{du}{dx}+p(x)\cdot y^n u= q(x)y^n$$ or equivalently $$\frac{du}{dx} +(1-n)p(x)u = (1-n)q(x)$$ as desired.