Solve the differential equation. 5\sqrt{xy}\frac{dy}{dx}=1, x, y>0

Cem Hayes 2021-09-18 Answered
Solve the differential equation.
\(\displaystyle{5}\sqrt{{{x}{y}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={1},{x},{y}{>}{0}\)

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Expert Answer

StrycharzT
Answered 2021-09-19 Author has 12588 answers
Step 1
Given,
\(\displaystyle{5}\sqrt{{{x}{y}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={1},{x},{y}{>}{0}\)
Step 2
Rewriting the equation,
\(\displaystyle{5}\sqrt{{{y}}}{\left.{d}{y}\right.}={\frac{{{1}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)
Step 3
Integrate both sides,
\(\displaystyle\Rightarrow{5}\int\sqrt{{{y}}}{\left.{d}{y}\right.}=\int{\frac{{{1}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}+{C}\), C is constant.
\(\displaystyle\Rightarrow{5}{\left({\frac{{{y}^{{{\frac{{{1}}}{{{2}}}}+{1}}}}}{{{\frac{{{1}}}{{{2}}}}+{1}}}}\right)}={\left({\frac{{{x}^{{-{\frac{{{1}}}{{{2}}}}+{1}}}}}{{-{\frac{{{1}}}{{{2}}}}+{1}}}}\right)}+{C}\)
\(\displaystyle\Rightarrow{5}{\left({\frac{{{2}{y}^{{{\frac{{{3}}}{{{2}}}}}}}}{{{3}}}}\right)}={2}{x}^{{{\frac{{{1}}}{{{2}}}}}}+{C}\)
\(\displaystyle\Rightarrow{y}^{{{\frac{{{3}}}{{{2}}}}}}={\frac{{{3}}}{{{10}}}}{\left({2}{x}^{{{\frac{{{1}}}{{{2}}}}}}+{C}\right)}\)
\(\displaystyle\Rightarrow{y}={\left({\frac{{{3}}}{{{10}}}}{\left({2}{x}^{{{\frac{{{1}}}{{{2}}}}+{C}}}\right)}\right)}^{{{\frac{{{2}}}{{{3}}}}}}\)
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