# Three questions in the resolution of the ODE $y'=\sqrt{y}$

Solve the differential equation $$y'=\sqrt{y}$$.

My attempt: it must be $$y(x)\ge 0$$. I notice that $$y_0 (x)=0$$ is a solution since $$y_0'(x)=0$$ and $$\sqrt{y_0(x)}=0$$.

Assuming then $$y(x)\ne0$$, that is $$y(x)>0$$, I can divide both sides of the equation and integrate in the interval $$[x_0,x]$$: $$\int_{x_0}^x \frac{y'(s)}{\sqrt{y(s)}}ds=\int_{x_0}^x ds$$ Hence $$2\sqrt{y(x)}-2\sqrt{y(x_0)}=x-x_0 \iff \sqrt{y(x)}=\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}$$ If the right hand side is negative the latter equations has no real solutions; hence assuming $$x \ge x_0-2\sqrt{y(x_0)}$$ it is $$\sqrt{y(x)}=\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}\iff y(x)=\left[\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}\right]^2$$ So the solutions are $$y_0(x)=0$$ or $$y(x)=\left[\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}\right]^2$$.

Questions:

1. I have consciously omitted any logical connective between the steps $$\frac{y'(x)}{\sqrt{y(x)}}=1$$ and $$\int_{x_0}^x \frac{y'(s)}{\sqrt{y(s)}}ds=\int_{x_0}^x ds$$ because I don't know what to use. I'm not sure if it is "$$f(x)=g(x)$$ for all $$x\in[a,b]$$ implies $$\int_a^b f(x)dx=\int_a^b g(x)dx$$" or "$$f(x)=g(x)$$ for all $$x \in[a,b]$$ if and only if $$\int_a^b f(x)dx=\int_a^b g(x)dx$$". I believe that it is implies, because, for example, $$f(x)=0$$ and $$g(x)=\sin x$$ have both integrals $$0$$ in the interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$ but of course they're not equal for all $$x\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$. What is the correct logical connective?

2. If the correct logical connective is "implies", shouldn't I check (as in algebraic equations) every time I integrate in ODEs that the solutions I find are actually solutions by substituting in the differential equation so that I'm not considering extra wrong solutions? I ask because I know that implication is, from a set theory point of view, an inclusion and so if at some point there is an implication I must check the reverse (because in solving equations we are searching for a set equivalence). Is this the case in ODEs?

3. My textbook adds the condition $$x \ge x_0-2\sqrt{y(x_0)}$$, I assume that this condition comes from the fact that, at some point, I have an equation between a square root and $$\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}$$ and so $$\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}$$ must be nonnegative; again coming back to logic, is this because the steps $$2\sqrt{y(x)}-2\sqrt{y(x_0)}=x-x_0$$ and $$\sqrt{y(x)}=\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}$$ are logically equivalent only applying that condition while, before isolating $$\sqrt{y(x)}$$, the quantity $$2\sqrt{y(x)}-2\sqrt{y(x_0)}$$ can be negative depending on $$x$$ and $$x_0$$ and so that condition of nonnegativity is not strictly necessary?

• It is not true that $$\int_a^bf(x)\,\mathrm{d}x=\int_a^bg(x)\,\mathrm{d}x$$ is equivalent to $$\forall{x\in[a,b]},f(x)=g(x).$$ However, it is true that that $$\forall{x\geq{x_0}},\int_{x_0}^xf(s)\,\mathrm{d}s=\int_{x_0}^xg(s)\,\mathrm{d}s$$ if and only if $$\forall{x\geq{x_0}},f(x)=g(x).$$ Nov 30 '21 at 13:59
• If the connective is implies, then yes, solution verification is necessary. If the connective is equivalence, then solution verification is unnecessary. Nov 30 '21 at 14:01
• The condition of nonnegativity is strictly necessary. Yes, $2\sqrt{y(x)}-2\sqrt{y(x_0)}$ can be negative, but regardless of the fact that this is true, it is necessarily the case that $2\sqrt{y(x)}$ is nonnegative, so whatever it is equal to, must also be nonnegative. Since $2\sqrt{y(x)}\geq0$ and $2\sqrt{y(x)}=x+2\sqrt{y(x_0)}-x_0,$ it is necessarily the case that $x+2\sqrt{y(x_0)}-x_0\geq0,$ which is the case if and only if $x\geq{x_0-2\sqrt{y(x_0)}}.$ There is simply no way around this. Nov 30 '21 at 14:05
• Also, I would like to add that even though $x\geq{x_0-2\sqrt{y(x_0)}}$ is a requirement for $$y(x)=\left[\frac{x}2+\sqrt{y(x_0)}-\frac{x_0}2\right]^2$$ to be a solution, it can be the case that $\forall{x\leq{x_0-2\sqrt{y(x_0)}}},y(x)=0,$ and this would give a solution on $\mathbb{R}$ that is everywhere differentiable. As such, this equation has two solutions on $\mathbb{R}:$ the piecewise solution I just constructed, which is once differentiable everywhere, but not twice differentiable everywhere; and the solution with $\forall{x\in\mathbb{R}},y(x)=0.$ Nov 30 '21 at 14:10