Solve the following initial value problems: xe^ y x -y + x dy dx … - Vidyadip

Question

Solve the following initial value problems:
$\text{xe}^{\frac{\text{y}}{\text{x}}}-\text{y + x}\frac{\text{dy}}{\text{dx}}=0,\text{y(e)}=0$

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Answer

$\text{xe}^{\frac{\text{y}}{\text{x}}}-\text{y + x}\frac{\text{dy}}{\text{dx}}=0,\text{y(e)}=0$
This is also a homogeneous equation.
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
$\text{xe}^{\text{v}}-\text{vx + x}\Big(\text{v + x}\frac{\text{dv}}{\text{dx}}\Big)=0$
$\text{xe}^{\text{v}}-\text{vx + xv}+\text{x}^2\frac{\text{dv}}{\text{dx}}=0$
$\text{xe}^{\text{v}}+\text{x}^2\frac{\text{dv}}{\text{dx}}=0$
$\text{e}^{\text{v}}=-\text{x}\frac{\text{dv}}{\text{dx}}$
$\frac{\text{dx}}{\text{x}}=-\frac{1}{\text{e}^{\text{v}}}\text{dv}$
On integrating both sides we get,
$\int\frac{\text{dx}}{\text{x}}=-\int\frac{1}{\text{e}^{\text{v}}}\text{dv}$
$\log_{\text{e}}\text{x}=-\int\text{e}^{-\text{v}}\text{dv}$
$\Rightarrow\ \log_{\text{e}}\text{x}=\text{e}^{-\frac{\text{y}}{\text{x}}}+\text{C}$ $(\because\ \text{y}=\text{vx})$
As given y(e) = 0
$\log_{\text{e}}\text{e}=\text{e}^{-\frac{0}{\text{e}}}+\text{C}$
$1=1+\text{C}$
$\Rightarrow\ \text{C}=0$
$\therefore \log_{\text{e}}\text{x}=\text{e}^{-\frac{\text{y}}{\text{x}}}$

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