Give the correct answer and solve the given equation: displaystyle{x}{y}{left.{d}{x}right.}-{left({y}+{2}right)}{left.{d}{y}right.}={0}

Question
Differential equations
asked 2020-12-05
Give the correct answer and solve the given equation:
\(\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({y}+{2}\right)}{\left.{d}{y}\right.}={0}\)

Answers (1)

2020-12-06
We have to find the solution for the given differential equation
\(\displaystyle{a}{y}{\left.{d}{x}\right.}={\left({y}+{2}\right)}{\left.{d}{y}\right.}.\)
We can separate the variables.
So,
\(\displaystyle{x}{y}{\left.{d}{x}\right.}={\left({y}+{2}\right)}{\left.{d}{y}\right.}\)
\(\displaystyle\Rightarrow{\left(\frac{{{y}+{2}}}{{y}}\right)}{\left.{d}{y}\right.}={x}{\left.{d}{x}\right.}\)
Integrating both sides we get,
\(\displaystyle\int{\left(\frac{{{y}+{2}}}{{y}}\right)}{\left.{d}{y}\right.}\int{x}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow\int{\left({1}+\frac{2}{{y}}\right)}{\left.{d}{y}\right.}=\int{x}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{\left.{d}{y}\right.}+\int{\left(\frac{2}{{y}}\right)}{\left.{d}{y}\right.}=\int{x}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow\int{\left.{d}{y}\right.}+{2}\int\frac{{\left.{d}{y}\right.}}{{y}}=\int{x}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{y}+{2} \log{{\left({y}\right)}}=\frac{{{x}^{2}}}{{2}}+{C}\), where C is a constant of integration.
Hence, the solution of the given differential equation is,
\(\displaystyle{y}+{2} \log{{\left({y}\right)}}=\frac{{{x}^{2}}}{{2}}+{C}\), C is a constant
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