How to find $u(x,y)$ given $\frac{\partial u}{\partial x}$?

Question

In the third line from the bottom, they find $u$, but I don't understand how they did it, could someone explain in more detail?

Answer 1

Partially differentiate the third line from the bottom ($u=\dots$) w.r.t to $x$ and you see that you get the line above it.

The constant $c_1$ is just the usual integration constant (that always disappears when you take the derivative) and $f(y)$, which is some function of $y$ (but not $x$), disappears because it doesn't depend explicitly on $x$, so to the eyes of the partial derivative, it too looks like a constant.

Finally, you have to know that $u=u(x,y)$ and no other variables. If, for instance, $u=u(x,y,z)$, then we would have to add yet another term $u=\dots+f(z)$.

Edit: As OP points out, $u=ux^2+\dots\;$ should indeed be $u=yx^2+\dots$

Answer 2

hint

You integrate according to $x$ . $y$ is like a constant.

for example, if

$$\frac{\partial u}{\partial x}=6x$$then

$$u(x,y)=3x^2+f(y)$$

and if $$\frac{\partial u}{\partial x}=6xy$$ then $$u(x,y)=3x^2y+f(y)$$