Using an Integrating Factor to solve differential equation

Question

I am trying to solve $$ (x^2y^3+y)dx+(x^3y^2-x)dy=0 $$ First of all we check this equation for exactness

$\frac{\partial P}{\partial y}=3y^2x^2+1$

$\frac{\partial Q}{\partial x}=3x^2y^2-1$

The partial derivatives are not equal to each other. Therefore, this equation is not exact.

Then I calculate the difference of the derivatives: $\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}=2$

Now I try to use the integrating factor in the form $z=xy$

$\frac{\partial z}{\partial x}=y, \frac{\partial z}{\partial y}=x$

Then:

$Q\frac{\partial z}{\partial x}-P\frac{\partial z}{\partial y}=y(x^3y^2-x)-x(x^2y^3+y)=-2xy$

and hence I get

$\frac{1}{\mu}\frac{\partial \mu}{\partial z}=\frac{\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}}{Q\frac{\partial z}{\partial x}-P\frac{\partial z}{\partial y}}=\frac{2}{-2xy}=-\frac{1}{xy}$

Now I can find the integrating factor by integrating the last equation and I get that $\mu=-xy$

Multiplying both sides of the first equation with $-xy$ I get

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

And when I calculate $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$ they are still not equal to each other, so the equation is still not exact and I can't solve it. What am I doing wrong?

Answer 1

Let us solve the ODE: $$(x^2y^3+y)dx+(x^3y^2-x)dy=0~~~~(1)$$ The oP did a mistake the integrating factor should be $\mu=\frac{1}{xy}$, then multiplying (1) by $\mu$ we get $$\left (xy^2+\frac{1}{x} \right) dx+ \left(x^2y-\frac{1}{y} \right)dy=0$$ Now this exact ODE. Which can be rearranged as $$xy(ydx+xdy)+\frac{dx}{x}-\frac{dy}{y}=0 \implies\int xy d(xy)+\int \frac{dx}{x}-\int\frac{dy}{y}=C $$ $$\implies (xy)^2/2+\ln x-\ln y=C \implies x^2y^2+2\ln (x/y)=D$$

Answer 2

In this type of exercise it is easy to see how it was constructed. Just group the terms by equal degree structure $$ [x^2y^3\,dx+x^3y^2\,dy] + [y\,dx-x\,dy]=0 $$ and remember that $d(x^ay^b)=x^{a-1}y^{b-1}[ay\,dx+bx\,dy]$ to identify $$ (xy)^2d(xy)-x^{-2}d(\frac yx)=0. $$ Now with new variables $u=xy$, $v=\frac yx$ one gets a separable equation $$ u^2\,du-\frac{v}{u}\,dv=0. $$ So indeed, multiplication with $u$ results in a completely integrable expression.

Answer 3

You multiplied by $-xy$ but that 's not the integrating factor. Which is $-\dfrac 1 {xy}$: $$\dfrac 1 {-xy}(x^2y^3+y)dx+ \dfrac 1 {(-xy)}(x^3y^2-x)dy=0$$ $$(-xy^2-\frac 1 x)dx+(-x^2y+\frac 1y)dy=0$$ It's exact now since you have: $$Pdx+Qdy=0$$ $$\partial_y P=-2xy \text { and } \partial_yQ=-2xy$$