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

Homework Help: Differential/Integration equation manipulation

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data

    a) Solve the differential equation
    dy/dx = x(y2+3)/y

    b) Find the unique function y(x) satisfying the differential equation with initial
    condition
    dy/dx = x2y, y(1) = 1

    2. Relevant equations


    3. The attempt at a solution
    With question a) I am no entirely sure but I have done
    dy/dx = x(y2+3)/y
    Let u = y2+3/y
    So dy/dx = x.u
    Integral of (dy/dx) = Integral (x.u)
    y = x2u2/2
    Then sub it back in. I'm not entirely sure


    With question b) I have done
    dy/dx = x2y
    dy . 1/y = x2dx
    integral of(1/y dx) = x3/3 + c
    Ln y = x3 /3 + c

    I am not sure if I am on the right track,
    I would appreciate any help.
    Thanks
     
  2. jcsd
  3. May 12, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    For a), try putting all the 'y's by the dy and all the x's by the dx like you did in part b

    Part b is correct. Now just sub in the condition y(1)=1
     
  4. May 12, 2010 #3

    lanedance

    User Avatar
    Homework Helper

    as rock points out, they're both separable
     
  5. May 12, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    This is equivalent to
    [tex]\frac{y}{y^2+ 3} dy= x dx[/tex]
    Integrate both sides.

    No, [itex]\int (dy/dx) dx= \int x u dx[/itex] but u is a function of y which is a function of x itself so you really have [itex]\int x f(x)dx[/itex] which cannot be integrated since you don't know "f(x)". Instead separate, as I suggested, and integrate the function of y with respect to y, the function of x with respect to x.

    Yes, what you have done for this is correct. Now, use the condition that when x= 1, y= 1 to solve ln(1)= 0= 13/2+ c for c.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook