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Simple integral

  1. Jan 27, 2017 #1
    1. The problem statement, all variables and given/known data
    181.png 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. jcsd
  3. Jan 27, 2017 #2

    haruspex

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    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.)
     
  4. Jan 27, 2017 #3

    LCKurtz

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    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.
     
  5. Jan 28, 2017 #4
    With y>0:
    $$\frac{dy}{dx}=\frac{x}{y}~\rightarrow~y^2=x^2+C$$
    $$\left\{ \begin{array}{lr} \mbox{for}~x>0 & y=x \\ \mbox{for}~x<0 & y=-x \end{array}\right.$$
     
  6. Jan 28, 2017 #5

    LCKurtz

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    Technically, without skipping steps you have$$
    \frac {y^2} 2 = \frac {x^2} 2 + C$$And you are ignoring the ##C## in your answer.
     
  7. Jan 28, 2017 #6
    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?
     
  8. Jan 28, 2017 #7

    LCKurtz

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    I figured you were using a shortcut, but still you have to solve ##y^2 = x^2 + C## for ##y##.
     
  9. Jan 30, 2017 #8
    $$y^2=x^2+C~\rightarrow~y=\pm \sqrt{x^2+C}$$
    This isn't a function since each x has two y's, so, with the help of y>0:
    $$y=+\sqrt{x^2+C}~\rightarrow~y=\pm (x+C)$$
    Snap1.jpg
     
  10. Jan 30, 2017 #9

    Ray Vickson

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    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.
     
  11. Jan 30, 2017 #10

    haruspex

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    Think that through again.
     
  12. Jan 31, 2017 #11
    $$y=+\sqrt{x^2+C}$$
    Snap1.jpg This is the final answer, i can't simplify further
     
  13. Jan 31, 2017 #12

    haruspex

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    Looks good. A possible quibble is that you are given y>0, not y≥0.
     
  14. Jan 31, 2017 #13
    Thank you LCKurtz, Ray Vickson and Haruspex
     
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