Does there exist a twice differentiable function $f:\mathbb R\to\mathbb R$ such that $f'(x)=f(x+1)-f(x)$ for all $x$ and $f''(0)\ne0$? Justify your answer.

Find all $c\in\mathbb R$ for which there exists an infinitely differentiable function $f:\mathbb R\to\mathbb R$ such that for all $n\in\mathbb N$ and $x\in\mathbb R$ we have $$f^{(n+1)}(x)>f^{(n)}(x)+c.$$

Find all solutions of the differential equation $zz" - 2z'z' = 0$ which pass through the point $x=1, z=1.$

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 $x=0.$

Assume that the system of differential equations $y'=-z^3$, $z'=y^3$ with the initial conditions $y(0)=1$, $z(0)=0$ has a unique solution $y=f(x)$, $z=g(x)$ defined for real $x$. Prove that there exists a positive constant $L$ such that for all real $x$, $$f(x+L)=f(x),\enspace g(x+L)=g(x).$$

Provided that the roots of the polynom $ X^n+a_1X^{n-1} +a_2X^{n-2} +\cdots +a_{n-1}X +a_n:\in\mathbb{R}[X] , $ of degree $ n\ge 2, $ are all real and pairwise distinct, prove that there exists is a neighbourhood $ \mathcal{V} $ of $ \left( a_1,a_2,\ldots ,a_n \right) $ in $ \mathbb{R}^n $ and $ n $ functions $ x_1,x_2,\ldots ,x_n\in\mathcal{C}^{\infty } \left( \mathcal{V} \right) $ whose values at $ \left( a_1,a_2,\ldots ,a_n \right) $ are roots of the mentioned polynom.

For the complex-valued function $f(x)$ which is continuous and absolutely integrable on $\mathbb{R}$, define the function $(Sf)(x)$ on $\mathbb{R}$: $(Sf)(x)=\int_{-\infty}^{+\infty}e^{2\pi iux}f(u)du$. (a) Find the expression for $S(\frac{1}{1+x^2})$ and $S(\frac{1}{(1+x^2)^2})$. (b) For any integer $k$, let $f_k(x)=(1+x^2)^{-1-k}$. Assume $k\geq 1$, find constant $c_1$, $c_2$ such that the function $y=(Sf_k)(x)$ satisfies the ODE with second order: $xy''+c_1y'+c_2xy=0$.

Let $u(t)$ be a continuous function in the system of differential equations $$\frac{dx}{dt} =-2y +u(t),\;\;\; \frac{dy}{dt}=-2x+ u(t).$$ Show that, regardless of the choice of $u(t)$, the solution of the system which satisfies $x=x_0 , y=y_0$ at $t=0$ will never pass through $(0, 0)$ unless $x_0 =y_0.$ When $x_0 =y_0 $, show that, for any positive value $t_0$ of $t$, it is possible to choose $u(t)$ so the solution is equal to $(0,0)$ when $t=t_0 .$

Let $f(x)$ be a bounded continuous function on $[0,+\infty)$. Prove that every solutions of the equation $y''+14y'+13y=f(x)$ are bounded continuous functions on $[0,+\infty)$

Find an integral formula for the solution of the differential equation $$\delta (\delta-1)(\delta-2) \cdots(\delta -n +1) y= f(x), \;\;\, x\geq 1,$$ for $y$ as a function of $f$ satisfying the initial conditions $y(1)=y'(1)=\ldots= y^{(n-1)}(1)=0$, where $f$ is continuous and $\delta$ is the differential operator $ x \frac{d}{dx}.$

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $$f(x)=f \left( \frac{x}{2} \right) + \frac{x}{2} f'(x), ~\forall x \in \mathbb{R}.$$ Prove that $f$ is a polynomial function of degree at most one. [hide=Note]The problem was posted quite a few times before: [url]https://artofproblemsolving.com/community/c7h100225p566080[/url] [url]https://artofproblemsolving.com/community/q11h564540p3300032[/url] [url]https://artofproblemsolving.com/community/c7h2605212p22490699[/url] [url]https://artofproblemsolving.com/community/c7h198927p1093788[/url] I'm reposting it just to have a more suitable statement for the [url=https://artofproblemsolving.com/community/c13_contests]Contest Collections[/url]. [/hide]

Let $g(t)$ and $h(t)$ be real, continuous functions for $t\geq 0.$ Show that any function $v(t)$ satisfying the differential inequality $$\frac{dv}{dt}+g(t)v \geq h(t),\;\; v(t)=c,$$ satisfies the further inequality $v(t)\geq u(t),$ where $$\frac{du}{dt}+g(t)u = h(t),\;\; u(t)=c.$$ From this, conclude that for sufficiently small $t>0,$ the solution of $$\frac{dv}{dt}+g(t)v = v^2 ,\;\; v(t)=c$$ may be written $$v=\max_{w(t)} \left( c e^{- \int_{0}^{t} |g(s)-2w(s)| \, ds} -\int_{0}^{t} e^{-\int_{0}^{t} |g(s')-2w(s')| \, ds'} w(s)^{2} ds \right),$$ where the maximum is over all continuous functions $w(t)$ defined over some $t$-interval $[0,t_0 ].$

Let $y_1$ and $y_2$ be two linearly independent solutions of the equation $$y''+P(x)y'+Q(x)=0.$$ Find the differential equation satisfied by the product $z=y_1 y_2$.

Prove that if the family of integral curves of the differential equation $$ \frac{dy}{dx} +p(x) y= q(x),$$ where $p(x) q(x) \ne 0$, is cut by the line $x=k$ the tangents at the points of intersection are concurrent.

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 $y''+\frac{1}{y^{3}}=f(x)y$.

Consider the function $y(x)$ satisfying the differential equation $y'' = -(1+\sqrt{x})y$ with $y(0)=1$ and $y'(0)=0.$ Prove that $y(x)$ vanishes exactly once on the interval $0< x< \pi \slash 2,$ and find a positive lower bound for the zero.

Let $x: [0, \infty) \to\Bbb R$ be a differentiable function. Prove that if for all t>1 $$x'(t)=-x^3(t)+\frac{t-1}{t}x^3(t-1)$$ then $\lim_{t\to\infty} x(t) = 0$

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).

$u(t)$ is solution of the following initial value problem. $$\begin{cases} u''(t) + u'(t) = \sin u(t) &\;\;(t>0),\\ u(0)=1,\;\; u'(0)=0 & \end{cases}$$ (1) Show that $u(t)$ and $u'(t)$ are bounded on $t>0$. (2) Find $\lim\limits_{t\to\infty} u(t)$ with proof.

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 $$y'+Ap(x)y=Br(x).$$

We want to find functions $p(t)$, $q(t)$, $f(t)$ such that (a) $p$ and $q$ are continuous functions on the open interval $(0,\pi)$. (b) $f$ is an infinitely differentiable nonzero function on the whole real line $(-\infty,\infty)$ such that $f(0)=f'(0)=f''(0)$. (c) $y=\sin t$ and $y=f(t)$ are solutions of the differential equation $y''+p(t)y'+q(t)y=0$ on $(0,\pi)$. Is this possible? Either prove this is not possible, or show this is possible by providing an explicit example of such $f,p,q$.

Show that if a connected, nowhere zero sectional curvature of Riemannian manifold, where symmetric (1,1)-tensor of the Levi-Civita connection covariant derivative vanishes, then the tensor is constant times the unit tensor. (translated by j___d)

A pond has been dug at the Indian Statistical Institute as an inverted truncated pyramid with a square base (see figure below). The depth of the pond is 6m. The square at the bottom has side length 2m and the top square has side length 8m. Water is filled in at a rate of $\tfrac{19}{3}$ cubic meters per hour. At what rate is the water level rising exactly $1$ hour after the water started to fill the pond? [img]https://cdn.artofproblemsolving.com/attachments/0/9/ff8cac4bb4596ec6c030813da7e827e9a09dfd.png[/img]

Show that if the differential equation $$M(x,y)\, dx +N(x,y) \, dy =0$$ is both homogeneous and exact, then the solution $y=y(x)$ satisfies that $xM(x,y)+yN(x,y)$ is constant.

The curve $y=y(x)$ satisfies $y'(0)=1.$ It satisfies the differential equation $(x^2 +9)y'' +(x^2 +4)y=0.$ Show that it crosses the $x$-axis between $$x= \frac{3}{2} \pi \;\;\; \text{and} \;\;\; x= \sqrt{\frac{63}{53}} \pi.$$