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Solve the differential equation

  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.
     
    Last edited: Jan 27, 2006
  2. jcsd
  3. Jan 28, 2006 #2
    I'm in a rush, so answers are quite incomplete and tend to be misleading, here they go
    1) it seems suitable for y=vx, dy=xdv/dx + v substitution

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

    [tex]y(1-x^2)dx = x(x^2+1)dy[/tex]
    [tex](x^2-1)dx/x(x^2+1) = dy/y[/tex]
    simply integrate both sides.

    3) how about giving initial conditions? ;)
     
  4. Jan 28, 2006 #3
    Answer for 3)

    Hi! I think I have the answer for 3).
    Here it goes:

    y=c1sin(2x+2) (E.1)

    y=c1sin(2x+2) ---> y^2=[c1sin(2x+2)]^2 (E.2)

    y=c1sin(2x+2) ---> y'/2=c1cos(2x+2) ---> (y'/2)^2=[c1cos(2x+2)]^2 (E.3)

    By adding (E.2) and (E.3) you have:

    y^2 + (y'/2)^2=(c1)^2 because (cosk)^2 + (sink)^2 = 1

    From (E.1) c1=y/sin(2x+2)

    We finally write:

    y^2 + (y'/2)^2=[y/sin(2x+2)]^2
     
  5. Jan 28, 2006 #4
    for equation 3 shouldn't it be y'/2=4c1cos(2x+c2) sin(2x+c2) for the middle step, because you have to use the chain rule.
     
  6. Jan 29, 2006 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First, since this was clearly labled "homework help", I've moved it to the homework section!

     
  7. Jan 29, 2006 #6
    Thanks for the help. For question 3 I'm suppose to have a new equation with no c1 or c2 in it. In other words I'm suppose to solve for c1, then take the derivative of the original equation, substitute c1 in for that, then take the derivative of the new equation, then solve for c2. Here is an easy problem to demonstrate what I mean.

    [tex]y=c_1e^{2x}+c_2e^{-x}[/tex]

    [tex]y'=2c_1e^{2x}-c_2e^{-x}[/tex]

    [tex]y"=4c_1e^{2x}+c_2e^{-x}[/tex]

    [tex]y'+y=3c_1e^{2x}[/tex]

    [tex]y"+y'=6c_1e^{2x}[/tex]

    [tex]c_1=\frac{y'+y}{3e^{2x}}[/tex]

    [tex]y"+y'=2y'+2y[/tex]

    [tex]y"-y'-2y=0[/tex]

    See.....no c1's or c2's
     
    Last edited: Jan 29, 2006
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