Let $$u(x)$$ and $$v(x)$$ satisfy the differential equation $${{du} \over {dx}} + p\left( x \right)u = f\left( x \right)$$ and $${{dv} \over {dx}} + p\left( x \right)v = g\left( x \right),$$ where $$p(x) f(x)$$ and $$g(x)$$ are continuous functions. If $$u\left( {{x_1}} \right) > v\left( {{x_1}} \right)$$ for some $${{x_1}}$$ and $$f(x)>g(x)$$ for all $$x > {x_1},$$ prove that any point $$(x,y)$$ where $$x > {x_1},$$ does not satisfy the equations $$y=u(x)$$ and $$y=v(x)$$

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 $$(0,k).$$