The solution of dydx=12x−y2 is given by

dxdy=2x−y2 (This is a linear equation in x).The integrating factor is e−∫2dy=e−2y.So ddyxe−2y=−y2e−2yIntegrating, we havexe−2y=−y2e−2y−2−∫ye−2ydy+ Const =y22e−2y+ye−2y2−12∫e−2ydy+ Const =y22e−2y+y2e−2y+e−2y4+C1∴ x=C1e2y+y22+y2+14

The solution of dydx=12x−y2 is given by

dxdy=2x−y2 (This is a linear equation in x).The integrating factor is e−∫2dy=e−2y.So ddyxe−2y=−y2e−2yIntegrating, we havexe−2y=−y2e−2y−2−∫ye−2ydy+ Const =y22e−2y+ye−2y2−12∫e−2ydy+ Const =y22e−2y+y2e−2y+e−2y4+C1∴ x=C1e2y+y22+y2+14

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The solution of $\frac{d y}{d x} = \frac{1}{2 x - y^{2}}$ is given by

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$y = C e^{- 2 x} + \frac{1}{4} x^{2} + \frac{1}{2} x + \frac{1}{4}$

$x = C e^{- y} + \frac{1}{4} y^{2} + \frac{1}{4} y + \frac{1}{2}$

$x = C e^{y} + \frac{1}{4} y^{2} + y + \frac{1}{2}$

$x = C e^{2 y} + \frac{1}{2} y^{2} + \frac{1}{2} y + \frac{1}{4}$

answer is D.

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Detailed Solution

$\frac{d x}{d y} = 2 x - y^{2}$ (This is a linear equation in $x$ ).

The integrating factor is $e^{- \int 2 d y} = e^{- 2 y}$ .

So $\frac{d}{d y} (x e^{- 2 y}) = - y^{2} e^{- 2 y}$

Integrating, we have

$\begin{array}{l} x e^{- 2 y} = \frac{- y^{2} e^{- 2 y}}{- 2} - \int y e^{- 2 y} d y + Const \\ = \frac{y^{2}}{2} e^{- 2 y} + \frac{y e^{- 2 y}}{2} - \frac{1}{2} \int e^{- 2 y} d y + Const \\ = \frac{y^{2}}{2} e^{- 2 y} + \frac{y}{2} e^{- 2 y} + \frac{e^{- 2 y}}{4} + C_{1} \\ ∴ x = C_{1} e^{2 y} + \frac{y^{2}}{2} + \frac{y}{2} + \frac{1}{4} \end{array}$