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Exercise 6.3 · Q27

Q.Verify that xy=log⁡y+cxy=\log y+c is a solution of dydx=y21−xy\dfrac{dy}{dx}=\dfrac{y^2}{1-xy}.

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✓ Free question

Differentiate xy=log⁡y+cxy=\log y+c implicitly w.r.t. xx: y+xdydx=1ydydxy+x\dfrac{dy}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}. Collect the dydx\dfrac{dy}{dx} terms: y=dydx(1y−x)=dydx⋅1−xyyy=\dfrac{dy}{dx}\left(\dfrac{1}{y}-x\right)=\dfrac{dy}{dx}\cdot\dfrac{1-xy}{y}. So dydx=y21−xy\dfrac{dy}{dx}=\dfrac{y^2}{1-xy}, exactly the given differential equation.

✓Final answer

Verified: xy=log⁡y+cxy=\log y+c solves dydx=y21−xy\dfrac{dy}{dx}=\dfrac{y^2}{1-xy}

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