Is squaring both sides of a differential equations valid?

Question

In the equation $ \int \sqrt{y} dy= \sqrt{x} dx $ is it valid to square both sides to obtain $ \int y dy = \int x dx $

Answer 1

Assuming it is $$\int \sqrt{y}dy=\int\sqrt{x}dx$$

Yes you can, but this may increase the number of solutions you get upon solving that equation for some initial term, and a specific value of x.

When you do take the square, keep in mind that square will also work upon $dy$ and $dx$, like in your case:
squaring yields $$ \int y \cdot (dy)^2=\int x\cdot(dx)^2$$

This is useless in your question, squaring both sides will just increase its difficulty, without yielding anything useful.

So, for $$\int \sqrt{y} dy=\int \sqrt{x} dx$$ simply integrate both the sides, you get $$ \frac{2}{3} y^{\frac{3}{2}}= \frac{2}{3} x^{\frac{3}{2}}+c$$ $$ y^{\frac{3}{2}}=x^{\frac{3}{2}}+c'$$ as your equation.

Answer 2

In general no, $\int f(y) \,dy = \int g(x) \,dx$ does not imply that $\int f(y)^2 \,dy = \int g(x)^2 \,dx$.

We can see this readily in your example by integrating both equations and observing that we get different solution sets. (I recommend you try carrying this out yourself---I've put the solution sets in a spoiler.)

The general solution to $\sqrt{y} \,dy = \sqrt{x} \,dx$ is $y = (x^{3 / 2} + C)^{2 / 3}$, but the general solution to $y \,dy = x \,dx$ is $y(x) = \pm \sqrt{x^2 + C}$.