Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
✓
Answer
Here, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\Big\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\Big\}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}\Big\{\log\Big(\frac{\text{vx}}{\text{x}}\Big)+1\Big\}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v + v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v}$
$\int\frac{1}{\text{v}\log\text{v}}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\log\log\text{v}=\log|\text{x}|+\log\text{C}$
$\log\text{v}=\text{xC}$
$\log\frac{\text{y}}{\text{x}}=\text{xC}$
$\frac{\text{y}}{\text{x}}=\text{e}^{\text{xC}}$
$\text{y}=\text{xe}^{\text{xC}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\Big\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\Big\}$
It is a homogeneous equation.
Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{\text{x}}\Big\{\log\Big(\frac{\text{vx}}{\text{x}}\Big)+1\Big\}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v + v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v}$
$\int\frac{1}{\text{v}\log\text{v}}\text{dv}=\int\frac{\text{dx}}{\text{x}}$
$\log\log\text{v}=\log|\text{x}|+\log\text{C}$
$\log\text{v}=\text{xC}$
$\log\frac{\text{y}}{\text{x}}=\text{xC}$
$\frac{\text{y}}{\text{x}}=\text{e}^{\text{xC}}$
$\text{y}=\text{xe}^{\text{xC}}$
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