Let $I\subset \mathbb{R}$ be an interval and $f(x)$ a continuous real-valued function on $I$. Let $y_1$ and $y_2$ be linearly independent solutions of $y''=f(x)y$ taking positive values on $I$. Show that for some positive number $k$ the function $k\cdot\sqrt{y_1 y_2}$ is a solution of $y''+\frac{1}{y^{3}}=f(x)y$.

A particle moves in $3$-space according to the equations: $$ \frac{dx}{dt} =yz,\; \frac{dy}{dt} =xz,\; \frac{dz}{dt}= xy.$$ Show that: (a) If two of $x(0), y(0), z(0)$ equal $0,$ then the particle never moves. (b) If $x(0)=y(0)=1, z(0)=0,$ then the solution is $$ x(t)= \sec t ,\; y(t) =\sec t ,\; z(t)= \tan t;$$ whereas if $x(0)=y(0)=1, z(0)=-1,$ then $$ x(t) =\frac{1}{t+1} ,\; y(t)=\frac{1}{t+1}, z(t)=- \frac{1}{t+1}.$$ (c) If at least two of the values $x(0), y(0), z(0)$ are different from zero, then either the particle moves to infinity at some finite time in the future, or it came from infinity at some finite time in the past (a point $(x, y, z)$ in $3$-space "moves to infinity" if its distance from the origin approaches infinity).

Under the assumption that the following set of relations has a unique solution for $u(t),$ determine it. $$ \frac{d u(t) }{dt} = u(t) + \int_{0}^{t} u(s)\, ds, \;\;\; u(0)=1.$$

Assume that the differential equation $$y'''+p(x)y''+q(x)y'+r(x)y=0$$has solutions $y_1(x)$, $y_2(x)$, $y_3(x)$ on the real line such that $$y_1(x)^2+y_2(x)^2+y_3(x)^2=1$$for all real $x$. Let $$f(x)=y_1'(x)^2+y_2'(x)^2+y_3'(x)^2.$$Find constants $A$ and $B$ such that $f(x)$ is a solution to the differential equation $$y'+Ap(x)y=Br(x).$$

Solve the equations $$ \frac{dy}{dx}=z(y+z)^n, \;\; \; \frac{dz}{dx} = y(y+z)^n,$$ given the initial conditions $y=1$ and $z=0$ when $x=0.$