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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 .

Can someone please help? Thanks in advance

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  • $\begingroup$ Express $y$ in terms of $u$ and subsitute in the initial equation. $\endgroup$ – Yves Daoust Nov 1 '16 at 8:52
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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.

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