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I have the following equation

$(xy^2 + x)dx + (yx^2 + y)dy=0$ and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.

Let $M = xy^2 + x $ and $N = yx^2 + y$

$My = 2xy \text{ and } Nx = 2xy $

$ \int M.dx ==> \int xy^2 + x = x^2y^2 + (x^2)/2 + g(y)$ \text{Partial of } (x^2y^2 + (x^2)/2 + g(y)) => xy^2 + g(y)' g(y)' = y $g(y) = y^2/2$ the general solution then is $C = x^2y^2/2 + x^2/2 + y^2/2$

Is this solution the same I would get if I had taken the Separate Equations route?

  • 1
    In particular, answers produced by programs such as Wolfram Alpha can be very unlike answers produced by human beings.2011-07-11

3 Answers 3

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Basically, yes. You would write $ \int \frac{y\, dy}{y^2 + 1} = \int - \frac{x \, dx}{x^2 + 1} $ $ \frac{1}{2} \ln(y^2 + 1) = - \frac{1}{2} \ln(x^2 + 1) + c $ Multiply both sides by 2 and exponentiate: $ y^2 + 1 = \frac{e^{2c}}{x^2+1} $ This is equivalent to your solution, with $C = (e^{2c}-1)/2$.

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    $y^2 + 1 = \frac{e^{2c}}{x^2+1} \Rightarrow (x^2+1)(y^2+1)= e^{2c}$. Do the multiplication on the left, divide by 2, and combine all the constants in a new one ;)2011-07-11
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Taking the other route you get y'y(x^2+1)=-xy^2-x. Denote $z=y^2$, hence z'=2yy'. So you get z'(x^2+1)=-2xz-x. Hence you have z'+\frac{2x}{x^2+1}z=-\frac{x}{x^2+1}. This gives you $z=\frac{k(x)}{x^2+1}$. Substituting, you have k'(x)=-x, which implies $k(x)=-\frac{x^2}{2}+C$. Hence $z=-\frac{x^2}{2(x^2+1)}+\frac{C}{x^2+1}$. Finally, $y^2=z$.
From here you get $2y^2(x^2+1)+x^2=C$, which is the answer you got.

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We can also try it this way,

$(xy^2 + x)dx + (yx^2 + y)dy=0$ $xdx +ydy +xy^2dx+yx^2dy$ $\frac{1}{2}(2xdx+2ydy) + \frac{1}{2}(2xy^2dx+2yx^2dy)=0$

$\frac{1}{2}d(x^2+y^2) + \frac{1}{2}d(x^2y^2) =0$

$x^2+y^2+ x^2y^2 +c=0$

:)

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    Take the gradient: $\nabla(x^2+y^2+x^2 y^2)$ and you get $\left\{2x(y^2+1), 2y(x^2+1)\right\}$. The homework problems ask you to go in reverse ('invert' the gradient).2012-05-08