# Solve the Bernoulli Differential equations. xy′ - y = xy^5

Question
Differential equations
Solve the Bernoulli Differential equations. $$\displaystyle{x}{y}′-{y}={x}{y}^{{5}}$$

2021-02-12

Here the given differential equation is
$$\displaystyle{x}{y}^{'}-{y}={x}{y}^{{5}}$$
which also can be written as
$$\displaystyle{y}^{'}-\frac{{y}}{{x}}={y}^{{5}}$$.
Let $$\displaystyle{v}={y}^{{-{{4}}}}{t}{h}{e}{n}\ {d}\frac{{v}}{{\left.{d}{x}\right.}}=-{4}{y}^{{-{{5}}}}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$$.The above De reduces to
$$\displaystyle-\frac{{y}^{{5}}}{{4}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}-\frac{{y}}{{x}}={y}^{{5}}$$
$$\displaystyle\Rightarrow-\frac{{1}}{{49}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}-\frac{{y}^{{-{{4}}}}}{{x}}={1}$$
$$\displaystyle\Rightarrow\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+\frac{{{4}{v}}}{{x}}=-{4}$$
$$\displaystyle\Rightarrow{x}^{{4}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+{4}{v}{x}^{{3}}=-{4}{x}^{{4}}$$
$$\displaystyle\Rightarrow\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({v}{x}^{{4}}\right)}=-{4}{x}^{{4}}$$
$$\displaystyle\Rightarrow{v}{x}^{{4}}=-{4}\int{x}^{{4}}{\left.{d}{x}\right.}+{c}$$
$$\displaystyle\Rightarrow{v}{x}^{{4}}=-\frac{{{4}{x}^{{5}}}}{{5}}+{c}$$
$$\displaystyle\Rightarrow{v}=-\frac{{{4}{x}}}{{5}}+{c}{x}^{{-{{4}}}}$$
$$\displaystyle\Rightarrow{y}^{{-{{4}}}}=-\frac{{{4}{x}}}{{5}}+{c}{x}^{{-{{4}}}}$$
$$\displaystyle\Rightarrow{y}={\left(-\frac{{{4}{x}}}{{5}}+{c}{x}^{{-{{4}}}}\right)}^{{4}}$$
where c is a arbitrary constant.

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$$\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{2}}+{1}\right)}$$
$$\displaystyle{\left({a}-{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$