Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrate dy/dx

  1. Aug 27, 2009 #1
    can some one explain to me how the set of all solutions for dy/dx = 3y
    y= Ce^3x
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 27, 2009 #2


    User Avatar
    Homework Helper

    Simply separate the variables.
    \frac{dy}{dx} & = 3y \\
    \frac 1 y \frac dy dx & = 3 \\
    \int \left(\frac 1 y\right) \left(\frac{dy}{dx}\right) \, dx & =\int 3 \, dx

    You should be able to finish from here.
  4. Aug 27, 2009 #3


    User Avatar
    Homework Helper

    Try rearranging your equation to get it in the form f(y)dy=g(x) dx. Then integrate both sides.

    This type of differential equation technique is called 'separation of variables'
  5. Aug 27, 2009 #4
    i got y= e^3x + e^c

    how does that become y= Ce^3x
  6. Aug 27, 2009 #5


    User Avatar
    Insights Author

    Staff: Mentor

    Instead of e3x + eC, you should have gotten e3x + C = e3xeC = C' e3x

    (Here, C' = eC. After all, eC is just a constant.)
  7. Aug 27, 2009 #6
    o ok thanks got it.
  8. Aug 27, 2009 #7
    this is variable separable. c is constant, then you can make it In c so that when using the e function, In c become only c.
  9. Aug 27, 2009 #8


    User Avatar
    Insights Author

    Staff: Mentor

    The function is LN, not IN. The letters come from Latin: logarithmus naturalis.
  10. Aug 27, 2009 #9
    Here's a proof of the general case that I got from Courant's Introduction to Calculus and Analysis. This is one of my favourite proofs:

    "If a function y = f(x) satisfies an equation of the form [tex] y' = \alpha y [/tex] where [tex] \alpha [/tex] is a constant, then y has the form [tex] y = f(x) = ce^{\alpha x} [/tex] where c is also a consant; conversely, every function of the form [tex] ce^{\alpha x} [/tex] satisfies the equation [tex] y' = \alpha y [/tex].

    It is clear that [tex] y = ce^{\alpha x} [/tex] satisfies this equation for any arbitrary constant c. Conversely, no other function satisfies the differential equation [tex] y' - \alpha y = = [/tex]. For if y is such a function, we consider the function [tex] u = ye^{-\alpha x} [/tex]. We then have

    [tex] u' = y'e^{-\alpha x} - \alpha y e^{-\alpha x} = e^{-\alpha x}(y' - \alpha y}) .[/tex]

    However, the right-hand side vanishes, since we have assumed that [tex] y' = \alpha y [/tex]; hence [tex] u' = 0 [/tex] so that u is a constant c and [tex] y = ce^{\alpha x} [/tex] as we wished to prove."
  11. Aug 28, 2009 #10
    yes its ln but I type In. Sorry.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?