Problem 2.10.6Test the following DE for exactness. If it is not exact, try to find an integrating factor. After the equation is made exact, solve by looking for integrable combinations. If you cannot find any, use the standard exact method.

$\displaystyle (x^{2}-y^{2}-y)\,dx+(y^{2}-x^{2}+x)\,dy=0.$

Answer.

We test for exactness. Is

$\displaystyle -2y-1=1-2x?$

No. We try to find an integrating factor.

1.Based on the symmetry of the problem, I'm going to guess a function $\displaystyle h=h(xy)$. Then we have

$\displaystyle h(xy)(x^{2}-y^{2}-y)\,dx+h(xy)(x+y^{2}-x^{2})\,dy=0.$

The exactness condition implies

$\displaystyle h'(xy)(x)(x^{2}-y^{2}-y)+h(xy)(-2y-1)=h'(xy)(y)(x+y^{2}-x^{2})+h(xy)(1-2x)\quad\Rightarrow$

$\displaystyle h'(xy)(x^{3}-xy^{2}-xy-xy-y^{3}+x^{2}y)=h(xy)(1-2x+1+2y)\quad\Rightarrow$

$\displaystyle h'(xy)(x^{3}-y^{3}-2xy-xy^{2}+x^{2}y)=h(xy)(2-2x+2y).$

I'm not sure this is going anywhere, because we'd have to write all the non-$\displaystyle h$ stuff as a function of $\displaystyle xy$, which I don't think is going to happen.

2.Let's try again, this time with a function $\displaystyle h=h(y/x)$. The DE is

$\displaystyle h\!\left(\dfrac{y}{x}\right)(x^{2}-y^{2}-y)\,dx+h\!\left(\dfrac{y}{x}\right)(x+y^{2}-x^{2})\,dy=0.$

The exactness condition yields

$\displaystyle h'\!\left(\dfrac{y}{x}\right)\left(\dfrac{1}{x}\ri ght)(x^{2}-y^{2}-y)+h\!\left(\dfrac{y}{x}\right)(-2y-1)=

h'\!\left(\dfrac{y}{x}\right)\left(-\dfrac{y}{x^{2}}\right)(x+y^{2}-x^{2})+h\!\left(\dfrac{y}{x}\right)(1-2x)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{y}{x}\right)\left(x-\dfrac{y^{2}}{x}-\dfrac{y}{x}\right)+h\!\left(\dfrac{y}{x}\right)(-2y-1)=

h'\!\left(\dfrac{y}{x}\right)(-\dfrac{y}{x}-\dfrac{y^{3}}{x^{2}}+y)+h\!\left(\dfrac{y}{x}\righ t)(1-2x)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{y}{x}\right)\left(x-\dfrac{y^{2}}{x}-\dfrac{y}{x}+\dfrac{y}{x}+\dfrac{y^{3}}{x^{2}}-y\right)=

h\!\left(\dfrac{y}{x}\right)(1-2x+1+2y)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{y}{x}\right)\left(x-\dfrac{y^{2}}{x}+\dfrac{y^{3}}{x^{2}}-y\right)=

h\!\left(\dfrac{y}{x}\right)(2-2x+2y)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{y}{x}\right)\left(1-\dfrac{y^{2}}{x^{2}}+\dfrac{y^{3}}{x^{3}}-\dfrac{y}{x}\right)=

h\!\left(\dfrac{y}{x}\right)\left(\dfrac{2}{x}-2+2\,\dfrac{y}{x}\right).$

Again, I'm not seeing how the non-$\displaystyle h$ stuff can be written as a function of $\displaystyle y/x$.

3.Let's go for a simpler one: $\displaystyle h=h(x)$. Then we have

$\displaystyle h(x)(x^{2}-y^{2}-y)\,dx+h(x)(x+y^{2}-x^{2})\,dy=0.$

The exactness condition implies

$\displaystyle h(x)(-2y-1)=h'(x)(x+y^{2}-x^{2})+h(x)(1-2x)\quad\Rightarrow$

$\displaystyle h(x)(-2y-1-1+2x)=h'(x)(x+y^{2}-x^{2}).$

Again, no prospect of writing it the way I want. By symmetry, I don't think a function of the type $\displaystyle h=h(y)$ will work.

4.Let's try the $\displaystyle h=h(x/y)$ type:

$\displaystyle h\!\left(\dfrac{x}{y}\right)(x^{2}-y^{2}-y)\,dx+h\!\left(\dfrac{x}{y}\right)(x+y^{2}-x^{2})\,dy=0.$

Exactness implies

$\displaystyle h'\!\left(\dfrac{x}{y}\right)\left(-\dfrac{x}{y^{2}}\right)(x^{2}-y^{2}-y)+h\!\left(\dfrac{x}{y}\right)(-2y-1)=

h'\!\left(\dfrac{x}{y}\right)\left(\dfrac{1}{y}\ri ght)(x+y^{2}-x^{2})+h\!\left(\dfrac{x}{y}\right)(1-2x)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{x}{y}\right)\left(-\dfrac{x^{3}}{y^{2}}+x+\dfrac{x}{y}\right)+h\!\lef t(\dfrac{x}{y}\right)(-2y-1-1+2x)=

h'\!\left(\dfrac{x}{y}\right)\left(\dfrac{x}{y}+y-\dfrac{x^{2}}{y}\right)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{x}{y}\right)\left(x+\dfrac{x}{y}-\dfrac{x^{3}}{y^{2}}-\dfrac{x}{y}-y+\dfrac{x^{2}}{y}\right)=

h\!\left(\dfrac{x}{y}\right)\left(2+2y-2x\right)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{x}{y}\right)\left(x-\dfrac{x^{3}}{y^{2}}-y+\dfrac{x^{2}}{y}\right)=

h\!\left(\dfrac{x}{y}\right)\left(2+2y-2x\right)\quad\Rightarrow$

$\displaystyle h'\!\left(\dfrac{x}{y}\right)\left(\dfrac{x}{y}-\dfrac{x^{3}}{y^{3}}-1+\dfrac{x^{2}}{y^{2}}\right)=

h\!\left(\dfrac{x}{y}\right)\left(\dfrac{2}{y}+2-2\,\dfrac{x}{y}\right).$

Still not seeing how this could work.

Does anyone see a mistake in my computations anywhere? See any new ideas?

Thanks!