Solution of differential equation x2d y-d x-y2 e xy-x-y-2 yx-y be given byA- xx-y-x logC ex-1B- xx-y-x logC ex-1C- xx-y-y logC ex-1D- xx-y-y logC ex-1

The correct option is D x(x – y) = y log (Ce^x – 1) x^2dy/dx 2y(x y)+y^2e^xe^ x^2/y =0 x^2/y^2dy/dx 2(x y)/y= e^xe^ x^2/y e^x^2/yx^2/y^2dy/dx 2x/ye^x^2/y+e^x= ...

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Byju's Answer

Standard XII

Mathematics

Solving Homogeneous Differential Equations

Solution of d...

Question

Solution of differential equation $x^{2} \frac{d y}{d x} + y^{2} e \frac{x (y - x)}{y} = 2 y (x - y)$ be given by

$x (x + y) = y l o g (C e^{x} - 1)$

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$x (x - y) = x l o g (C e^{x} - 1)$

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$x (x + y) = x l o g (C e^{x} - 1)$

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$x (x - y) = y l o g (C e^{x} - 1)$

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Solution

The correct option is D
$x (x - y) = y l o g (C e^{x} - 1)$

$x^{2} \frac{d y}{d x} - 2 y (x - y) + y^{2} e^{x} e^{\frac{- x^{2}}{y}} = 0 \frac{x^{2}}{y^{2}} \frac{d y}{d x} - \frac{2 (x - y)}{y} = - e^{x} e^{\frac{- x^{2}}{y}} e^{\frac{x^{2}}{y}} \frac{x^{2}}{y^{2}} \frac{d y}{d x} - \frac{2 x}{y} e^{\frac{x^{2}}{y}} + e^{x} = 0$
Let $e^{\frac{x^{2}}{y}} = t e^{\frac{x^{2}}{y}} (\frac{2 x y - x^{2} \frac{d y}{d x}}{y^{2}}) = \frac{d t}{d x} \Rightarrow e^{\frac{x^{2}}{y}} \frac{2 x}{y} - \frac{x^{2}}{y^{2}} e^{\frac{- x^{2}}{y}} \frac{d y}{d x} = \frac{d t}{d x} - \frac{d t}{d x} + 2 t + e^{x} = 0 \Rightarrow \frac{d t}{d x} - 2 t = e^{x}$
On solving, $t = - e^{x} + C e^{2 x}$
Where C is constant of integration
or $e^{\frac{x^{2}}{y}} = - 1 + C E^{x}$
$(\frac{x^{2}}{y} - x) = l o g (C e^{x} - 1) x (x - y) = y l o g (C e^{c} - 1)$

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Solution of differential equation x2dydx+y2ex(y−x)y=2y(x−y) be given by

Solution of differential equation $x^{2} \frac{d y}{d x} + y^{2} e \frac{x (y - x)}{y} = 2 y (x - y)$ be given by