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:

The solution of the differential equation $3 x y' - 3 y + (x^2 - y^2)^{1/2} = 0$,satisfying the condition $y(1) = 1$ is

$I: y^{\prime}=\frac{y+x}{x} ; \quad II: y^{\prime}=\frac{x^2+y}{x^3} ; \quad III: y^{\prime}=\frac{2xy}{y^2-x^2}$
$S1$: 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.

The general solution of the differential equation $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$ is

The general solution of the differential equation $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$ is

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