Differential equations with five scenarios It is my first week doing Differential Equations, and I am stuck at the following question:
Let $f$ be a function with a continuous derivative at any point in $\mathbf R$. Let $c$ be a real number. It is given that $u(x)$ is a solution to the differential equation $y'=f(y)$.
Which claims are correct?
a. function $v(x)=u(x+c)$ is a solution to $y'=f(y)$.
b. if $u'(c)=0$, then $u(x)$ is a constant function.
c. if $u(x)$ is not a constant function, then it is monotonous (either rising or falling).
d. if $f(v)$ is also a solution to $y'=f(y)$ and satisfies $u(c)<v(c)$, then $u(x)<v(x)$ for every $x$ for which both are defined.
e. the general solution of $y'=f(y)$ is $y=u(x+C)$.
I am pretty sure that a. and e. are correct. I suspect that b. is incorrect and that d. is correct, but I am far from sure. I have no idea regarding c. What is throwing me off in particular is that $u$ is a function of $x$, whereas $y'$ is a function of $y$, and I am not sure re the implications of that.
Any help would be amazing!
 A: a.- True. By a direct computation, taking the derivative of $v$ we obtain $v'(x)=u'(x+c)$ for any $c\in\mathbb{R}$. Thus, using that $u(x)$ solves the ODE we obtain $$
v'(x)=u'(x+c)=f(u(x+c))=f(v(x)).
$$
Hence $v$ is a solution of the ODE.
b.- True. This is an application of the Cauchy-Lipschitz Theorem (or "existence and uniqueness theorem"). By contradiction assume you have a solution $u(x)$ such that $u'(c)=0$ but $u(x)$ is not identically constant. Then, define $c^\star:=u(c)$. Thus, defining the constant function $$\tilde{u}(x)\equiv c^\star=u(c) \quad \text{for all }x\in\mathbb{R},$$ we found a solution of the ODE. In fact, differentiating $\tilde u$ we obtain $$
\forall x\in\mathbb{R}, \qquad \tilde u'(x)=0=f(c^\star).
$$
But then, both $u$ and $\tilde u$ are solutions and satisfy$$
u(c)=\tilde u(c)=c^\star.
$$
Thus, by the Cauchy-Lipschitz Theorem $u\equiv \tilde u$, and thus $u(x)$ is constant. Contradiction!
c.- True. This is an application of the Cauchy-Lipschitz Theorem (or "existence and uniqueness theorem") and goes exactly as the previous question. By contradiction assume you have a non-monotonic solution. Since the solution change monotonicity, there exists $x^\star\in\mathbb{R}$ such that $$
u'(x^\star)=0 \quad \implies \quad f(u(x^\star))=f(c^\star)=0.
$$
Now, define $c^\star:=u(x^\star)$. Thus, defining the constant function $\tilde{u}(x)\equiv c^\star$ for all $x\in\mathbb{R}$, we found a solution of the ODE, since $$
\forall x\in\mathbb{R}, \qquad \tilde u'=f(c^\star)=0.
$$
But then, both solutions satisfy$$
u(x^\star)=\tilde u(x^\star)=c^\star.
$$
Thus, by the Cauchy-Lipschitz Theorem $u\equiv \tilde u$, and thus $u(x)$ is constant. Contradiction!
d.- True. An easy way to prove this is by contradiction using the Cauchy-Lipschitz Theorem (the existence and uniqueness theorem). Assume that it is false, that is, there exists $\tilde{c}$ such that $$
u(\tilde c)\geq v(\tilde{c}).
$$
Then, there exists $x^\star\in[c,\tilde{c}]$ (without loss of generality you can assume $\tilde{c}>c$) such that $$
u(x^\star)=v(x^\star).
$$
Thus, by the Cauchy-Lipschitz Theorem $u\equiv v$, which is a contradiction with the hypothesis $u(c)<v(c)$.
I let you to solve the last question. Good luck :)
