Integrate dy/dx

  1. can some one explain to me how the set of all solutions for dy/dx = 3y
    is.
    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. statdad

    statdad 1,479
    Homework Helper

    Simply separate the variables.
    [tex]
    \begin{align*}
    \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
    \end{align*}
    [/tex]

    You should be able to finish from here.
     
  4. rock.freak667

    rock.freak667 6,228
    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. i got y= e^3x + e^c

    how does that become y= Ce^3x
     
  6. Mark44

    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. o ok thanks got it.
     
  8. 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. Mark44

    Staff: Mentor

    The function is LN, not IN. The letters come from Latin: logarithmus naturalis.
     
  10. 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. yes its ln but I type In. Sorry.
     
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