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 $y(x)=\left[\frac{x}{2}+\sqrt{y(x_0)}-\frac{x_0}{2}\right]^2$.


  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?

  • $\begingroup$ 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 $$\forall{x\geq{x_0}},f(x)=g(x).$$ $\endgroup$
    – Angel
    Nov 30 '21 at 13:59
  • $\begingroup$ If the connective is implies, then yes, solution verification is necessary. If the connective is equivalence, then solution verification is unnecessary. $\endgroup$
    – Angel
    Nov 30 '21 at 14:01
  • $\begingroup$ 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. $\endgroup$
    – Angel
    Nov 30 '21 at 14:05
  • 1
    $\begingroup$ 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 $\forall{x\in\mathbb{R}},y(x)=0.$ $\endgroup$
    – Angel
    Nov 30 '21 at 14:10

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