Solve the differential equation: (x^ 2 + 3xy + y^2) dx - x^ 2 dy … - Vidyadip

Question

Solve the differential equation:
$(x^{2} + 3xy + y^2) dx - x^{2} dy = 0,\text{given that } y = 0, \text{when } x = 1.$

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Answer

$\text{(x}^{2} + \text{3xy + y}^{2}) \text{ dx }{- }\text{x}^{2} \text{ dy} = 0,$
$\frac{\text{dy}}{\text{dx}} = \frac{\text{x}^{2} + \text{3xy + y}^{2}}{\text{x}^{2}}$
$\text{let y = vx}$
$\frac{\text{dy}}{\text{dx}} = \text{v + x}\frac{\text{dv}}{\text{dx}}$
$\therefore\text{v + x}\frac{\text{dv}}{\text{dx}} = 1 + \text{3v + v}^{2}$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}} = \text{v}^{2} + \text{2v + 1}$
$\Rightarrow\frac{\text{dv}}{\text{(v + 1)}^{2}} = \frac{\text{dx}}{\text{x}}$
Integrating both sides
$\Rightarrow -\frac{1}{\text{v + 1}} =\log|\text{x}|+\text{C}$
$\Rightarrow \frac{{-}\text{x}}{\text{x + y}} =\log|\text{x}|+\text{C}$
$\text{When x = 1, y = 0}\Rightarrow\text{C} = {-1}$
$\Rightarrow \frac{{-}\text{x}}{\text{x + y}} =\log|\text{x}|-1$
$\Rightarrow\text{y = (x + y)}\log\text{|x|}$
or $\text{y} = \frac{\text{x}\log|\text{x}|}{1 - \log |\text{x}|}$

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