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solve the following differential equation.

$\tag 1(x^2-2x+2y^2)\,dx+2xy\,dy=0$


dividing (1) throughout by $y^2$ we have,

$\tag 2 \left(\frac{x^2}{y^2}+2-2\frac{x}{y^2}\right)dx+2(\frac{x}{y})dy=0$

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You can use the \tag command to tag equations. – Pedro Tamaroff Feb 14 '13 at 13:14
What does equation (1) mean? the LHS is a non-zero 1-form, and the RHS is $0$! – Mercy King Feb 14 '13 at 13:15
There must be some typos in the formulas. – Student Feb 14 '13 at 13:15
Multiply equation $(1)$ by $x$. This will yield an exact differential equation. The method here was used to find the integrating factor $x$. – David Mitra Feb 14 '13 at 13:18
@Julien a 1-form $\omega=a(x,y)dx+b(x,y)dy$ is zero iff $a=b=0$. – Mercy King Feb 14 '13 at 14:10
up vote 4 down vote accepted

Illustrating @DavidMitra's idea: multiply through by $x$:


We want to find a function $F(x,y)$ such that

$$\frac{\partial F}{\partial x} = x^3-2x^2+2xy^2$$

$$\frac{\partial F}{\partial y} = 2x^2y$$

From the former equation, we integrate with respect to $x$ and get

$$F(x,y) = \frac{1}{4} x^4 - \frac{2}{3} x^3 + x^2 y^2 + g(y)$$

From the latter, integrate with respect to $y$ and get:

$$F(x,y) = x^2 y^2 + f(x)$$

Comparing the two equations, we see that $f(x) = \frac{1}{4} x^4 - \frac{2}{3} x^3 $ and $g(y)=0$. Therefore

$$F(x,y) = \frac{1}{4} x^4 - \frac{2}{3} x^3 + x^2 y^2$$

Solutions of your differential equation satisfy $F(x,y) = C$, a constant.

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Perfect, +1. In view of @Mecy's comments, do you think we can really call this a differential equation? As is the question properly stated? – 1015 Feb 14 '13 at 16:32
Sure, why not? Divide through by $dx$ and you can see it easier. What I showed is a standard technique, where $F$ is known as the first integral of the equation. – Ron Gordon Feb 14 '13 at 16:35
Because what is given is a differential $1$-form. Dividing by $dx$ does not make sense in this context. And, as far as I know, you don't divide differential forms. Like @Mercy says, such an object is zero if and only if its coordinates are both zero, which leads to a system of equations which have nothing to do with differential equations (well, they're order zero diff eq, ok). – 1015 Feb 14 '13 at 16:42
Dividing by $dx$ never really makes sense in any context, but it is a convenient mnemonic. In this case, it reveals the equivalent diff eq'n. Think about solving that diff eq'n: you would "multiply" by "$dx$ and integrate both sides to reveal something hopefully integrable. What I showed here merely systemizes this. – Ron Gordon Feb 14 '13 at 16:48
Ok, well, we can't agree on everything. In my view, the question is not stated properly... – 1015 Feb 14 '13 at 16:58


In view of @Mercy's comments, I interpret your question as: find all constant functions $F(x,y)$ such that $$ dF=(x^2-2x+2y^2)dx+2xydy. $$

Nervertheless, I think you maybe meant your question to be: solve the ODE $$ x^2-2x+2y^2+2xyy'=0. $$

Setting $z=y^2$, we have $z'=2yy'$ so the equation becomes $$ xz'+2z=2x-x^2. $$ This is linear of first order.

A general method for solving this is the integrating factor method:

As pointed out by David Mitra, in this case, this is very easy since this leads to $$ x^2z'+2xz=2x^2-x^3\quad\Leftrightarrow\quad (x^2z)'=2x^2-x^3. $$

Then integrate the latter.

Once you have $z$, take $\pm\sqrt{z}$ where it is nonnegative to find $y$.

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