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

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

Question
Differential equations
asked 2021-02-11
Solve the Bernoulli Differential equations. \(\displaystyle{x}{y}′-{y}={x}{y}^{{5}}\)

Answers (1)

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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