ordinary differential equations

Question

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

Answer 1

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.

Answer 2

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

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

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

Let $y=xu$,

Then $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

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

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

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

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

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

This belongs to an Abel equation of the second kind

Check whether this ODE satisfy the special case in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=4:

$(3u^3+4u)(2(-2)+(4u)')=(3u^3+4u)(-4+4)=0$

$4u(-3u^2+(3u^3+4u)')=4u(-3u^2+9u^2+4)\neq0$

$\therefore$ not satisfy the special case in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=4

Let $v=x+\dfrac{3u^2+4}{4}$,

Then $x=v-\dfrac{3u^2+4}{4}$

$\dfrac{dx}{du}=\dfrac{dv}{du}-\dfrac{3u}{2}$

$\therefore4uv\left(\dfrac{dv}{du}-\dfrac{3u}{2}\right)=-2\left(v-\dfrac{3u^2+4}{4}\right)^2-3u^2\left(v-\dfrac{3u^2+4}{4}\right)$

$4uv\dfrac{dv}{du}-6u^2v=-2v^2+(3u^2+4)v-\dfrac{9u^4+24u^2+16}{8}-3u^2v+\dfrac{9u^4+12u^2}{4}$

$4uv\dfrac{dv}{du}=-2v^2+(6u^2+4)v+\dfrac{9u^4-16}{8}$

$v\dfrac{dv}{du}=-\dfrac{v^2}{2u}+\dfrac{(3u^2+2)v}{2u}+\dfrac{9u^4-16}{32u}$

In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $v=\dfrac{1}{w}$,

Then $\dfrac{dv}{du}=-\dfrac{1}{w^2}\dfrac{dw}{du}$

$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=-\dfrac{1}{2uw^2}+\dfrac{3u^2+2}{2uw}+\dfrac{9u^4-16}{32u}$

$\dfrac{dw}{du}=-\dfrac{(9u^4-16)w^3}{32u}-\dfrac{(3u^2+2)w^2}{2u}+\dfrac{w}{2u}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2