# ordinary differential equations

test the exactness of the O.D.E $(4xy+2x^2 y)dx+(2x^3+3y^2)dy=0$ and hence find the potential function which is the general solution.I tried to solve it and I reached ending up failing to get the integrating factor.please help me

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Does the close-form expression exist for this ODE? – Jack Apr 13 '12 at 7:16
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Sep 10 '12 at 1:45

Consider $$I(x,y)dx + J(x,y)dy = 0.$$ If this is an exact ODE we should be able to find a potential function $F(x,y)$ such that $$\frac{\partial}{\partial x}F(x,y) = I(x,y) = 4xy + 2x^2y$$ and $$\frac{\partial}{\partial y}F(x,y) = J(x,y) = 2x^3 + 3y^2.$$ We also know by Clairaut's theorem that the mixed partial derivatives should be equal, i.e. $$\frac{\partial^2}{\partial x \partial y} F(x,y) = \frac{\partial^2}{\partial y \partial x} F(x,y).$$ Equivalently, $$\frac{\partial}{\partial y} I(x,y) = \frac{\partial}{\partial x} J(x,y).$$ This is very useful, since it makes it easy to check whether or not an ODE is exact. $$\frac{\partial}{\partial y} I(x,y) = 4x + 2x^2$$ $$\frac{\partial}{\partial x} J(x,y) = 6x^2$$ These two expressions are not equal, so we conclude that this is not an exact ODE.

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OK, that shows it's not exact, but then OP asks for an integrating factor. Got one? – Gerry Myerson Apr 15 '12 at 9:38
@wolfram;Thanks for help.Still I can't solve it, may you give me more help on that? – Mwalyaje Apr 20 '12 at 18:30
+1 for cool nick – leo May 23 '12 at 4:04

Hint:

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

$(2x^3+3y^2)dy=-((2x^2+4x)y)dx$

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

Let $u=y^2$ ,

Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$

$\therefore\dfrac{(2x^3+3y^2)}{2y}\dfrac{du}{dx}=-(2x^2+4x)y$

$(2x^3+3y^2)\dfrac{du}{dx}=-(4x^2+8x)y^2$

$(2x^3+3u)\dfrac{du}{dx}=-(4x^2+8x)u$

This belongs to an Abel equation of the second kind.

Let $v=u+\dfrac{2x^3}{3}$ ,

Then $u=v-\dfrac{2x^3}{3}$

$\dfrac{du}{dx}=\dfrac{dv}{dx}-2x^2$

$\therefore3v\left(\dfrac{dv}{dx}-2x^2\right)=-(4x^2+8x)\left(v-\dfrac{2x^3}{3}\right)$

$3v\dfrac{dv}{dx}-6x^2v=-(4x^2+8x)v+\dfrac{8x^4(x+2)}{3}$

$3v\dfrac{dv}{dx}=(2x^2-8x)v+\dfrac{8x^4(x+2)}{3}$

$v\dfrac{dv}{dx}=\dfrac{(2(x-2)^2-8)v}{3}+\dfrac{8x^4(x+2)}{9}$

Let $s=x-2$ ,

Then $\dfrac{dv}{dx}=\dfrac{dv}{ds}\dfrac{ds}{dx}=\dfrac{dv}{ds}$

$\therefore v\dfrac{dv}{ds}=\dfrac{(2s^2-8)v}{3}+\dfrac{8(s+2)^4(s+4)}{9}$

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