Determine the equation of the curve passing through the origin, in the form $$y=f(x),$$ which satisfies the differential equation $${{dy} \over {dx}} = \sin \left( {10x + 6y} \right).\,$$

Let $$y=f(x)$$ be a curve passing through $$(1,1)$$ such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area $$2.$$ From the differential equation and determine all such possible curves.

A normal is drawn at a point $$P(x,y)$$ of a curve. It meets the $$x$$-axis at $$Q.$$ If $$PQ$$ is of constant length $$k,$$ then show that the differential equation describing such curves is $$y = {{dy} \over {dx}} = \pm \sqrt {{k^2} - {y^2}} $$

Find the equation of such a curve passing through $$(0,k).$$

If $$\left( {a + bx} \right){e^{y/x}} = x,$$ then prove that $${x^3}{{{d^2}y} \over {d{x^2}}} = {\left( {x{{dy} \over {dx}} - y} \right)^2}$$