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: Homework help

  1. Jan 27, 2006 #1
    I need someone to look over my work if possible

    1) Solve the differential equation
    xdy=(5y+x+1)dx

    Here is what I did:
    x=(5y+x+1)dx/dy
    [tex]x=5xy+\frac{x^2}{2}+x[/tex]
    [tex]0=5xy+\frac{x^2}{2}[/tex]
    [tex]\frac{-x^2}{2}=5xy[/tex]
    [tex]y=\frac{-x}{10}[/tex]

    2) Solve: [tex]y(x^2-1)dx+x(x^2+1)dy=0[/tex]

    Here is what I did:

    [tex]x^2ydx-ydx+x^3dy+xdy=0[/tex]

    [tex]\frac{xdy-ydx}{x^2}+ydx+xdy=0[/tex]

    [tex]d(\frac{y}{x})+d(xy)=0[/tex]
    This is where I am stuck at. I need to have something multiply the first differential that will yield something in the form of [tex]\frac{y}{x}[/tex] and also have the same thing multiply the second differential and will yield something in the form of xy.

    3) Find a differential equation with the solution [tex]y=c_1sin(2x+c_2)[/tex]

    Here is what I did:

    [tex]y'=2c_1cos(2x+c_2)[/tex]

    and since
    [tex]c_1=\frac{y}{sin(2x+c_2)}[/tex]

    [tex]y'=\frac{2ycos(2x+c_2)}{sin(2x+c_2}[/tex]

    [tex]y'=2ycot(2x+c_2)[/tex]

    I'm not sure what to do after this. I know I need to somehow get rid of the [tex]c_2[/tex] constant, but I don't know how to do this.
     
  2. jcsd
  3. Jan 27, 2006 #2
    In the second one just get all of the x and dx terms on one side and all of the y and dy terms on the other. That should only take one step.

    [tex]
    y\left( {x^2 - 1} \right)dx + x\left( {x^2 + 1} \right)dy = 0
    [/tex]

    [tex]
    \Rightarrow \left( {\frac{{x^2 - 1}}{{x^2 + 1}}} \right)dx = - \frac{1}{y}dy
    [/tex]

    Now integrate both sides.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook