If $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$,$x > 0$,$\phi > 0$,and $y(1) = -1$,then $\phi(y^2/4)$ is equal to:

$I: y^{\prime}=\frac{y+x}{x} ; \quad II: y^{\prime}=\frac{x^2+y}{x^3} ; \quad III: y^{\prime}=\frac{2xy}{y^2-x^2}$
$S1$: Differential equations given by $I$ and $II$ are homogeneous differential equations.
$S2$: Differential equations given by $II$ and $III$ are homogeneous differential equations.
$S3$: Differential equations given by $I$ and $III$ are homogeneous differential equations.

If $X = x + h, Y = y + k$ transforms $\frac{dy}{dx} = \frac{2x + 3y - 7}{3x + 2y - 8}$ to a homogeneous differential equation,then $(h, k) =$

Let the solution curve $y = y(x)$ of the differential equation $\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{dy}{dx} = x + \left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y$ pass through the points $(1, 0)$ and $(2\alpha, \alpha)$,where $\alpha > 0$. Then $\alpha$ is equal to:

The substitution $y = z^{\alpha}$ transforms the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$ into a homogeneous differential equation for

If the gradient of the tangent at any point $(x, y)$ of a curve which passes through the point $\left( 1, \frac{\pi}{4} \right)$ is $\left\{ \frac{y}{x} - \sin^2\left( \frac{y}{x} \right) \right\}$,then the equation of the curve is: