Question

❌❌❌Solve the following Differential Equation:- ❌❌❌
$xy .\frac{dy}{dx} - {y}^{2} = {(x + y)}^{2} . {e}^{ \frac{ - y}{x} }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\:xy\dfrac{dy}{dx} - {y}^{2} = {(x + y)}^{2} \bigg({e}\bigg)^{ \dfrac{ - y}{x} }$

can be rewritten as

$\rm :\longmapsto\:xy\dfrac{dy}{dx} = {(x + y)}^{2} \bigg({e}\bigg)^{ \dfrac{ - y}{x} } + {y}^{2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{{(x + y)}^{2} \bigg({e}\bigg)^{ \dfrac{ - y}{x} } + {y}^{2}}{xy}$

Since, degree of each term of numerator and denominator is same, so its a homogeneous differential equation.

So, we substitute

$\red{\rm :\longmapsto\:y = vx}$

So, above expression can be rewritten as

$\rm :\longmapsto\:\dfrac{d}{dx}(vx) = \dfrac{{(x + vx)}^{2} \bigg({e}\bigg)^{ \dfrac{ - vx}{x} } + {(vx)}^{2}}{x(vx)}$

$\rm :\longmapsto\:v\dfrac{d}{dx}x + x\dfrac{d}{dx}v = \dfrac{ {x}^{2} {(1 + v)}^{2} {e}^{ - v} + {x}^{2} {v}^{2} }{ {vx}^{2} }$

$\rm :\longmapsto\:v + x\dfrac{dv}{dx} = \dfrac{ {x}^{2} \bigg[{(1 + v)}^{2} {e}^{ - v}+{v}^{2} \bigg]}{ {vx}^{2} }$

$\rm :\longmapsto\:v + x\dfrac{dv}{dx} = \dfrac{ {(1 + v)}^{2} {e}^{ - v}+{v}^{2} }{ v}$

$\rm :\longmapsto\:v + x\dfrac{dv}{dx} = \dfrac{ {(1 + v)}^{2} {e}^{ - v}}{ v} + v$

$\rm :\longmapsto\:x\dfrac{dv}{dx} = \dfrac{ {(1 + v)}^{2} {e}^{ - v}}{ v}$

On separating the variables, we get

$\rm :\longmapsto\:\dfrac{v{e}^{v}}{ {(v + 1)}^{2} } \: dv \: = \: \dfrac{dx}{x}$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{v{e}^{v}}{ {(v + 1)}^{2} } \: dv \: = \: \displaystyle\int\rm \dfrac{dx}{x}$

can be rewritten as

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{(v + 1 - 1){e}^{v}}{ {(v + 1)}^{2} } \: dv \: = \: logx + c$

$\rm :\longmapsto\:\displaystyle\int\rm \bigg[\dfrac{v + 1}{ {(v + 1)}^{2} } - \dfrac{1}{ {(v + 1)}^{2} } \bigg]{e}^{v} \: dx = logx + c$

$\rm :\longmapsto\:\displaystyle\int\rm \bigg[\dfrac{1}{ {(v + 1)} } - \dfrac{1}{ {(v + 1)}^{2} } \bigg]{e}^{v} \: dx = logx + c$

We know,

$\boxed{\tt{ \dfrac{d}{dx} \dfrac{1}{x} = - \dfrac{1}{ {x}^{2} }}}$

and

$\boxed{\tt{ \displaystyle\int\rm {e}^{x}\bigg[f(x) + f'(x)\bigg]dx = {e}^{x} \: f(x) + c}}$

So, using this, we get

$\rm :\longmapsto\:{e}^{v} \times \dfrac{1}{v + 1} = logx + c$

$\rm :\longmapsto\: \dfrac{{e}^{v}}{v + 1} = logx + c$

$\rm :\longmapsto\: \dfrac{{\bigg(e\bigg)}^{\dfrac{y}{x} }}{\dfrac{y}{x} + 1} = logx + c$

$\rm :\longmapsto\: \dfrac{{\bigg(e\bigg)}^{\dfrac{y}{x} }}{\dfrac{y + x}{x}} = logx + c$

$\rm :\longmapsto\: \dfrac{x{\bigg(e\bigg)}^{\dfrac{y}{x} }}{y + x} = logx + c$