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Question

If $$(x-y)e^{\dfrac {x}{x-y}}=a$$, prove that $$y\dfrac {dy}{dx}+x=2y$$.


Solution

$$(x - y){e^{\dfrac{x}{{x - y}}}} = a$$

Differentiate both sides w.r.t.$$x$$

$$\left( {1 - \dfrac{{dy}}{{dx}}} \right){e^{\dfrac{x}{{x - y}}}} + \left( {x - y} \right){e^{\dfrac{x}{{x - y}}}} \times \left[ {\dfrac{{x - y - x\left( {1 - \dfrac{{dy}}{{dx}}} \right)}}{{{{\left( {x - y} \right)}^2}}}} \right] = 0$$

$${e^{\dfrac{x}{{x - y}}}}\left[ {\left( {1 - \dfrac{{dy}}{{dx}}} \right) + \left( {x - y} \right)\left( {\dfrac{{x - y - x\left( {1 - \dfrac{{dy}}{{dx}}} \right)}}{{{{\left( {x - y} \right)}^2}}}} \right)} \right] = 0$$

$$\left( {1 - \dfrac{{dy}}{{dx}}} \right) + \left( {\dfrac{{x\dfrac{{dy}}{{dx}} - y}}{{\left( {x - y} \right)}}} \right) = 0$$

$$x - y - x\dfrac{{dy}}{{dx}} + y\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} - y = 0$$

$$y\dfrac{{dy}}{{dx}} + x = 2y$$


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