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

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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