Question

Solve. \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+\frac{{3}}{{x}}{y}={27}{y}^{{\frac{{1}}{{3}}}}{1}{n}{\left({x}\right)},{x}{>}{0}\)

Answer 1

Here the given differential equation is
\(\displaystyle{y}^{'}+\frac{{{3}{y}}}{{x}}={27}{y}^{{\frac{{1}}{{3}}}}\)
Let \(\displaystyle{v}={y}^{{\frac{{2}}{{3}}}}{t}{h}{e}{n}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}=\frac{{2}}{{3}}{y}^{{\frac{{1}}{{3}}}}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\).With respect to this transformation, the above DE reduces to \(\displaystyle\frac{{3}}{{2}}{y}^{{\frac{{1}}{{3}}}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+\frac{{{3}{y}}}{{x}}={27}{y}^{{\frac{{1}}{{3}}}}\)
\(\displaystyle\Rightarrow\frac{{3}}{{2}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+\frac{{{3}{y}^{{\frac{{2}}{{3}}}}}}{{x}}={27}\)
\(\displaystyle\Rightarrow\frac{{1}}{{2}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+\frac{{{y}^{{\frac{{2}}{{3}}}}}}{{x}}={9}\)
\(\displaystyle\Rightarrow\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+\frac{{{2}{y}^{{\frac{{2}}{{3}}}}}}{{x}}={18}\)
\(\displaystyle\Rightarrow\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+{2}\frac{{v}}{{x}}={18}\)
\(\displaystyle\Rightarrow{x}^{{2}}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}+{2}{x}{v}={18}{x}^{{2}}\)
\(\displaystyle\Rightarrow\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({x}^{{2}}{v}\right)}={18}{x}^{{2}}\)
\(\displaystyle\Rightarrow{x}^{{2}}{v}=\int{18}{x}^{{2}}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle\Rightarrow{x}^{{2}}{v}={6}{x}^{{3}}+{c}\)
\(\displaystyle\Rightarrow{v}={6}{x}+{c}{x}^{{-{{2}}}}\)
\(\displaystyle\Rightarrow{y}^{{\frac{{2}}{{3}}}}={6}{x}+{c}{x}^{{-{{2}}}}\)
\(\displaystyle\Rightarrow{y}={\left({6}{x}+{c}{x}^{{-{{2}}}}\right)}^{{\frac{{3}}{{2}}}}\)
Where c is arbitrary constant.