# How can i show that this differential equation is exact?

The Problem is to show (proof) a given differential equation is exact and to solve the equation.

$$(2y^2+6xy-x^2)\ dx\ +\ (y^2 + 4xy + 3x^2)\ dy = 0$$

My Approach: i can see this is ordinary differential equation, it should be non-linear and of first order. I thought i could solve the problem by integrating each part of the equation by self, using the other variable as a constant. this led to the following: $$(2y^2+6xy-x^2)\ dx\ = -\frac{x^3}{3}+3x^2y+2xy^2$$ $$(y^2 + 4xy + 3x^2)\ dy = 3x^2y+2xy^2+\frac{y^3}{3}$$ So alltogether i would get:

$$(2y^2+6xy-x^2)\ dx\ +\ (y^2 + 4xy + 3x^2)\ dy = -\frac{x^3}{3}+6x^4y+2xy^2+\frac{y^3}{3}$$

And i have: $$-\frac{x^3}{3}+6x^4y+2xy^2+\frac{y^3}{3}=0$$

But i that cannot be the solution, it seems to be to easy. And additional i don't have a clue to give the Proof, i am looking for. I think i am stuck.

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Hint: Take the derivative of the first expression in parens w.r.t. $y$ and the second in parens w.r.t. $x$. Are they equal? – Amzoti Jun 10 '13 at 20:48
What makes a Diff.eqn. exact? How do you solve exact diff.eqns? – gt6989b Jun 10 '13 at 20:51
its a differential equation with an existing $F(x,y)$ so: $\frac{\partial F(x,y)}{\partial x}=p(x,y)$ ... but i cannot make the connection needed, to solve the given equation. – Toralf Westström Jun 10 '13 at 21:14
equations 2,3,4 have only one differentials in them. they seem to be missing an integral sign. – Maesumi Jun 10 '13 at 22:03

Like you said: $$\frac{\partial F(x,y)}{\partial x} = p(x,y), \quad \text{and} \frac{\partial F(x,y)}{\partial y} = q(x,y),$$ then $F(x,y) = C$ solves $$p(x,y) + q(x,y)\frac{dy}{dx} = 0,$$ basically we take derivative w.r.t. $x$ in $F(x,y) = C$.
Now your equation reads: $$\frac{\partial F(x,y)}{\partial x} = 2y^2+6xy-x^2 \implies F(x,y) = \int ( 2y^2+6xy-x^2) dx + \color{red}{H(y)},$$ where $H(y)$ is the key, because a function purely based on $y$ is lost when we take $\partial_x$. Then $$F(x,y) = 2y^2x +3x^2y - \frac{x^3}{3}+ H(y).\tag{1}$$ Here we can use the second condition: $$\frac{\partial F(x,y)}{\partial y} = y^2 + 4xy + 3x^2,$$ plugging (1) into above: $$\frac{\partial }{\partial y}\left(2y^2x +3x^2y - \frac{x^3}{3}+ H(y)\right) = y^2 + 4xy + 3x^2,$$ for first three terms, take partial derivative w.r.t. $y$. For $H(y)$, which is a function only of $y$-variable, take $\partial/\partial y$ is the same as $d/dy$, hence this gives you (notice some terms get canceled): $$\frac{d }{d y} H(y) = \ldots$$ intergrating both w.r.t. $y$ as if you are integrating only a one independent variable ODE, then plugging back to (1) you will get your $F(x,y)$.
i think the answer could be like that: $$\frac{d }{d y} H(y) = - \frac{d}{d y}\left(2y^2x +3x^2y - \frac{x^3}{3}\right) + y^2 + 4xy + 3x^2$$ $$\frac{d }{d y} H(y) = - \left(3x^2 + 4xy \right) + y^2 + 4xy + 3x^2$$ $$\frac{d }{d y} H(y) = y^2+4xy-4xy+3x^2-3x^2 = y^2$$ $$\frac{d }{d y} H(y) = y^2$$ $$H(y)= \int y^2\ dy = \frac{1}{3}\cdot y^3= \frac{y^3}{3}$$ plugging $H(y)$ into (1): $$F(x,y) = 2y^2x +3x^2y - \frac{x^3}{3}+ H(y)$$ $$F(x,y) = 2y^2x +3x^2y - \frac{x^3}{3}+ \frac{y^3}{3}$$ Is that already the solution of the whole diff. equation? – Toralf Westström Jun 11 '13 at 7:41
@ToralfWestström The whole solution should be $F(x,y)= \text{Constant}$ then. – Shuhao Cao Jun 11 '13 at 12:32