# Solve the given differential equation. \frac{dy}{dx}=\frac{y}{x}

Solve the given differential equation.
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{y}}}{{{x}}}}$$

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Willie
Step 1
We have to solve the differential equation:
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{y}}}{{{x}}}}$$
Solving this equation by variable separable method(shifting same variable with same derivatives),
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{y}}}{{{x}}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}={\frac{{{\left.{d}{x}\right.}}}{{{x}}}}$$....(1)
Step 2
Integrating equation (1) both sides, we get
$$\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}=\int{\frac{{{\left.{d}{x}\right.}}}{{{x}}}}$$
$$\displaystyle{\ln{{\left({y}\right)}}}={\ln{{\left({x}\right)}}}+{\ln{{\left({c}\right)}}}$$
(Using formula $$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}={\ln{{\left({x}\right)}}}+{c}$$)
(Here we have used ln(c) as an arbitrary constant due to all terms in logarithmic)
Now, using the property of logarithmic $$\displaystyle{\log{{\left({a}\right)}}}+{\log{{\left({b}\right)}}}={\log{{\left({a}{b}\right)}}}\ {\quad\text{and}\quad}\ {\log{{\left({a}\right)}}}={\log{{\left({b}\right)}}}\Rightarrow{a}={b}$$, we get
$$\displaystyle{\ln{{\left({y}\right)}}}={\ln{{\left({x}\right)}}}+{\ln{{\left({c}\right)}}}$$
$$\displaystyle{\ln{{\left({y}\right)}}}={\ln{{\left({x}{c}\right)}}}$$
y=xc
$$\displaystyle{c}={\frac{{{y}}}{{{x}}}}$$
Hence, solution of differential given equation is $$\displaystyle{c}={\frac{{{y}}}{{{x}}}}$$.