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Solution to DiffEq

  1. Dec 16, 2013 #1
    I've come across this problem while self studying Ordinary Differential Equations and I really need help. The problem asks me to simply eliminate derivatives, I do not need to separate. The book shows the answer, but not the steps.

    [itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]


    So far, I'm here...

    [itex]\frac{dy}{dx}[/itex]= [itex]\frac{y^2}{1-xy}[/itex]
    y' = [itex]\frac{y^2}{1-xy}[/itex]
    y' = [itex]\frac{y}{\frac{1}{y}-x}[/itex]
    y = y'([itex]\frac{1}{y}[/itex]-x)
    y = [itex]\frac{y'}{y}[/itex] - y'x
    [itex]\frac{y'}{y}[/itex] = y+ y'x
    now integrating both sides we get..
    ln (y) = ∫(y + y'x dx)

    But how do I integrate y+y'x?
  2. jcsd
  3. Dec 16, 2013 #2


    Staff: Mentor

    CAn't you use the product rule x'y + y'x
  4. Dec 16, 2013 #3
    We want to integrate.
    xy must be the solution to ∫(y'x+y dx) because the derivative of xy = y'x+y(1). I just need to know how to integrate this -> ∫(y'x+y dx)
  5. Dec 16, 2013 #4
    ∫(y'x+y dx) is a non-sens.
    ∫(y'x+y) dx is correct.
    (y'x+y) = (xy)'
    ∫(y'x+y) dx = ∫(yx)' dx = ? (obvious)
  6. Dec 16, 2013 #5

    (y'x+y) = (xy)'
    Can you please explain
    Thank you!
  7. Dec 16, 2013 #6


    Staff: Mentor

    given two function f(x) and g(x) then the derivative of the product f*g is:

    (fg)' = f'g + fg' i.e. the product rule so you look at your integral notice that it fits the right hand side and rewrite it to be

    the indefinite integral: integral( (f'g+fg')dx) = integral( (fg)'dx) = fg
    Last edited: Dec 16, 2013
  8. Dec 16, 2013 #7
    Is this how you came to this conclusion?

    = xy'(1+y)
    = x([itex]\frac{d(1+y)}{dx}[/itex])
    = x(0 + y')
    = x(y')
    = xy'
    = (xy)'
    so, ∫(xy)' dx = xy
    ∫[itex]\frac{y'}{y}[/itex] dx = ∫(xy)' dx [itex]\rightarrow[/itex] ln(y) + C = xy
    Last edited: Dec 16, 2013
  9. Dec 16, 2013 #8


    Staff: Mentor

    I think you're over complicating this. Have you learned the product rule when you took calculus 1? Just by inspection y + y'x you can see that there are two functions x and y with x' =1 and y' = y'

    Hence integrating it just means you replace the y+y'x term with (xy)' then you can conclude that the integrated answer is simply y'x via the fundamental theorem of calculus right?

    So is your problem going from integral(xy)' to get the ln(y) ?
  10. Dec 16, 2013 #9
    How so? By using the product rule of differentation, we would be solving the equation backwards. I did not ask for the answer, I asked how to integrate y+y'x. It is clear that the derivative of xy is indeed (dy/dx)x+ y(1), but not every problem involving y in the integral will be as obvious as this problem. From the product rule of diff, sure y'(xy) = y'x+y, but I wanted to know how to integrate y'x+y, which would in fact be the way shown above, not over complicating it but simply using as a learning experience. Thanks for the help :-)
  11. Dec 22, 2013 #10
    Equation made exact.

    [itex]{\frac{dy}{dx} = \frac{y^2}{1-xy}}[/itex]
    [itex] (y^2)dx+(xy-1)dy= 0 [/itex]
    [itex] \frac{∂M}{∂y} = 2y [/itex]
    [itex] \frac{∂N}{∂x} = y [/itex]
    [itex] 2y ≠ y[/itex]
    [itex]h(y) = \frac{∂N/∂x-∂M/∂y}{M} = \frac{-1}{y}[/itex]
    [itex]\mu(y) = e^∫h(y) dy[/itex]
    [itex]\mu(y) = 1/y [/itex]
    [itex] \frac{1}{y}[(y^2)dx+(xy-1)= 0] [/itex]
    [itex] (y)dx+(x-\frac{1}{y})dy= 0 [/itex]
    [itex] \frac{∂M}{∂y}= 1 = \frac{∂N}{∂x} [/itex]
    [itex]\frac{∂f}{∂x} = y[/itex]
    [itex] f(x,y) = xy + \phi(y) [/itex]
    [itex] \frac{∂f}{∂y} = x + \phi'(y) [/itex]
    [itex]x- \frac{1}{y} = x + \phi'(y) [/itex]
    [itex] \phi'(y) = -1/y [/itex]
    [itex] \phi(y) = -ln(y) [/itex]
    [itex] f(x,y) = xy - ln y + d [/itex]
    [itex] xy - ln y + d = c [/itex]
    [itex] xy - ln y = \tilde{c} [/itex]
    [itex] xy - ln y [/itex] [itex]\equiv 0[/itex]
    [itex] ln y = xy [/itex]
    Last edited: Dec 22, 2013
  12. Dec 23, 2013 #11


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    How about substituting [itex]z=x y[/itex]? Then you get
    [tex]z'=y+x y'.[/tex]
    [tex]y'=\frac{z'-y}{x}=\frac{x z'-z}{x^2} \stackrel{!}{=} \frac{y^2}{1-xy}=\frac{z^2}{x^2} \frac{1}{1-z}.[/tex]
    After some algebra you find
    which is solvable by separation of variables:
    [tex]\int \mathrm{d} z \frac{1-z}{z}=\int \mathrm{d} x \frac{1}{x}[/tex]
    [tex]\ln |z|-z=\ln|x|+C \; \Rightarrow\; \ln \left |\frac{z}{x} \right| = z+C \; \Rightarrow \; \ln |y| = x y +C,[/tex]
    which shows that your (implicit) solution is indeed solving the equation with the special integration constant [itex]C=0[/itex] and restricted to [itex]y>0[/itex].
  13. Dec 23, 2013 #12
    Is this how you would go about the alegebra for the z-sub?
    [tex]y' = \frac{y^2}{1-xy} [/tex]
    [tex]z = xy \Rightarrow y' = \frac{z'-y}{x} [/tex]
    [tex]y'=\frac{xz'-z}{x^2} [/tex]
    [tex] \frac{y^2}{1-xy}=\frac{xz'-z}{x^2}[/tex]
    [tex] \frac{z^2}{x^2-zx^2}=\frac{xz'-z}{x^2}[/tex]
    [tex]\frac{z^2}{1-z}= xz' - z[/tex]
    [tex] \frac{z^2}{1-z} + z =\frac{xz'}{1}[/tex]
    [tex] \frac{1-z}{z}=\frac{1}{xz'}[/tex]
    [tex] \frac{1-z}{z} * z' =\frac{1}{x}[/tex]
    *Also, must an equation be homogeneous in order for z-substitution to work? Kinda confused because I don't see any homogeneity properties and this one's clearly solvable by making this substitution.
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