Simple integral

1. Jan 27, 2017

Karol

1. The problem statement, all variables and given/known data
why do i need the information that y>0?

2. Relevant equations
$$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$

3. The attempt at a solution
$$\frac{dy}{dx}=\frac{x}{y}~\rightarrow~\int y~dy=\int x~dx~\rightarrow~y^2=x^2~\rightarrow~y=x$$
I square root y2 and if y<0 it works also

2. Jan 27, 2017

haruspex

A couple of inaccuracies.
You should allow a constant of integration since you are not given any initial condition.
When taking the square root you must allow for opposite signs.
If your answer allows y>0 then your answer is wrong. You need to write it such that y is > 0. (Merely adding the text "y>0" would do, but maybe there's something better.)

3. Jan 27, 2017

LCKurtz

To add to haruspex's remarks, when you add the constant of integration and show your steps, I think it will be apparent where you use the $y>0$ information.

4. Jan 28, 2017

Karol

With y>0:
$$\frac{dy}{dx}=\frac{x}{y}~\rightarrow~y^2=x^2+C$$
$$\left\{ \begin{array}{lr} \mbox{for}~x>0 & y=x \\ \mbox{for}~x<0 & y=-x \end{array}\right.$$

5. Jan 28, 2017

LCKurtz

Technically, without skipping steps you have$$\frac {y^2} 2 = \frac {x^2} 2 + C$$And you are ignoring the $C$ in your answer.

6. Jan 28, 2017

Karol

Yes, right, i omitted the division by 2 for shortening. after it comes $~y^2=x^2+C$ and the rest, right?
The condition y>0 must have been imposed since without it $~y=\pm x$ and this isn't a function since every x has 2 y's. am i right?

7. Jan 28, 2017

LCKurtz

I figured you were using a shortcut, but still you have to solve $y^2 = x^2 + C$ for $y$.

8. Jan 30, 2017

Karol

$$y^2=x^2+C~\rightarrow~y=\pm \sqrt{x^2+C}$$
This isn't a function since each x has two y's, so, with the help of y>0:
$$y=+\sqrt{x^2+C}~\rightarrow~y=\pm (x+C)$$

9. Jan 30, 2017

Ray Vickson

If $C > 0$ the functions $y = \sqrt{x^2+C}$ and $y = + (x+C)$ or $y = -(x+C)$ are different; their graphs do not even look the same. One of them is differentiable everywhere, while the other two are not.

10. Jan 30, 2017

haruspex

Think that through again.

11. Jan 31, 2017

Karol

$$y=+\sqrt{x^2+C}$$
This is the final answer, i can't simplify further

12. Jan 31, 2017

haruspex

Looks good. A possible quibble is that you are given y>0, not y≥0.

13. Jan 31, 2017

Karol

Thank you LCKurtz, Ray Vickson and Haruspex