# Is it possible to solve a separable equation a different way and still arrive at the same answer?

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?

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If you solve the same problem in different ways, you should always expect the same answer, otherwise something is fishy. Sometimes surprises occur when the same answer is actually in a completelly different, non recognizable ways... Also keep in mind that sometimes, the constants/intervals for the solutions make some difference for integrals and differential equations... –  N. S. Jul 11 '11 at 18:30
In particular, answers produced by programs such as Wolfram Alpha can be very unlike answers produced by human beings. –  André Nicolas Jul 11 '11 at 19:18

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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Sorry, but I don't see the equivalence. Can you please show how? –  user10695 Jul 11 '11 at 18:30
$y^2 + 1 = \frac{e^{2c}}{x^2+1} \Rightarrow (x^2+1)(y^2+1)= e^{2c}$. Do the multiplication on the left, divide by 2, and combine all the constants in a new one ;) –  N. S. Jul 11 '11 at 18:33

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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Ahhhh gradients.. –  Alex Hirzel May 8 '12 at 14:30
is it what you call these in your country? we just call it the short cut method, or the smart method. :) –  Tomarinator May 8 '12 at 14:37
Some Calculus textbooks/instructors prescribe many inverse gradient problems similar to this. –  Alex Hirzel May 8 '12 at 14:42
Inverse Gradient problems, googled it, seems like a hybrid of differentiation and inverse, what is it actually? –  Tomarinator May 8 '12 at 14:50
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). –  Alex Hirzel May 8 '12 at 14:55