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$If y + x \frac{dy}{dx} = x \frac{φ (xy)}{φ^{1} (xy)} then φ (xy) is$

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Kex2/2

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Kexy/2

Kexy

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Correct option is A

Put xy=v⇒y+xdydx=dvdxdvdx=xφ(v)φ1(v)∫φ1(v)φ(v)dv=∫xdxln |φ(v)|=x22+ln cφ(v)=Kex22φ(xy)=Kex22

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If y+xdydx=xφ(xy)φ1(xy) then φ(xy) is

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$If y + x \frac{dy}{dx} = x \frac{φ (xy)}{φ^{1} (xy)} then φ (xy) is$