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Homework Help: Solutions to Differential equations

  1. Oct 7, 2012 #1
    I'm taking my first ODE course, and I'm unsure what is meant in this question when it asks "Solve the following differential equation" (I have a list of DEs to solve). Some of them are really messy and I can't figure out an implicit solution. Would an explicit solution be acceptable?


    [itex]\frac{dy}{dx} = \frac{x}{x^2y+y^3}[/itex]

    Since this isn't an exact equation, I transform it and find the integrating factor

    [itex]\frac{dy}{dx}(x^2y+y^3) - x = 0 [/itex]

    and setting the partials [itex]\frac{∂}{∂x}N(x,y)μ(x) = \frac{∂}{∂y}M(x,y)μ(x) [/itex]

    I get [itex]μ(x) = \frac{1}{x^2+y^2}[/itex]

    I thought the integrating factor was supposed to be only a function of x, I may have made a mistake there.

    Then I take [itex] \frac{∂}{∂y}\int M(x,y)μ(x) dx [/itex]

    and I find [itex]f(x,y)=C=\frac{x^2+y^2}{2} - ln(x^2+y^2)[/itex]

    This just doesn't look right, I feel like I'm using the wrong method. Can someone clear this up for me please?
  2. jcsd
  3. Oct 7, 2012 #2
    First convert it to:


    You can do that. I get:

    [tex]xdx-(x^2 y+y^3)dy=0[/tex]

    Now compute:

    [tex]\mu=\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)[/tex]

    I get [itex]\mu=2y[/itex] but I did that quick so double-check it. Therefore, the integrating factor is:

    [tex]e^{-\int \mu dy}[/tex]

    Now apply that integrating factor to the DE and solve it as an exact DE. And to "solve" a DE is to find a function y(x) for example, which when substituted into the DE, satisfies the expression.
  4. Oct 7, 2012 #3
    Yes that's right, but I was under the impression the integrating factor would be a function of x, seeing as when we assume ∂μ/∂y = 0, this would be false if μ = 2y, and consequently fail to make the equation exact. So can the integrating factor be a function of y, or even both x and y?
  5. Oct 8, 2012 #4
    Yes, integrating factor can be a function of x and y for example:

    [tex]y(x^3-y) dx-x(x^3+y)dy=0[/tex]

    the integrating factor is [itex]\frac{1}{yx^2}[/itex]

    Find "Differential Equations" by Rainville and Bedient. Good book I think.
  6. Oct 8, 2012 #5


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    Science Advisor

    Generally speaking, "implicit" solutions cannot be written as explicit functions of x. Whether your instructor will accept such solutions even when, by trying a little harder, you could have found an explicit function, you will have to ask your instructor!
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