Determining integrating factor

Question

Consider

$$2x^2y\ln(y)+(x^3+x)y'=0$$

I want to determine the integrating factor $m(x,y)$ of the form $m(x,y)=m(xy)$.

Let $a(x,y):=2x^2y\ln(y)$ and $b(x)=x^3+x$.

So I want to find $m(x,y)$ such that $(ma)_y=(mb)_x$ where $(\cdot)_x$ denotes $\frac{d(\cdot)}{dx}$.

$(ma)_y=m_y \cdot x \cdot (2x^2y\ln(y))+m \cdot (2x^2\ln(y)+2x^2)$

$(m_b)_x=m_x \cdot y \cdot (x^3+x)+m \cdot (3x^2+1)$

We used the chain rule there. $m(x,y)_y=m_y\cdot x$

Now, \begin{align}(ma)_y&=(m_b)_x \\ \Leftrightarrow m_y \cdot x \cdot (2x^2y\ln(y))+m \cdot (2x^2\ln(y)+2x^2)&=m_x \cdot y \cdot (x^3+x)+m \cdot (3x^2+1)\end{align}

Now what I don't understand is that they treat $m_y$ as equal to $m_x$, namely $m_y=m_x=m'$ and arrive at

$\Leftrightarrow m'\cdot(2x^3y\ln(y)-yx^3-yx)=m\cdot(3x^2+1-2x^2\ln(y)-2x^2)$

But perhaps this is wrong anyways.

I would appreciate it if somemone could show me how this is done. Also different or easier/faster methods are fine

Answer 1

$$2x^2y\ln(y)+(x^3+x)y'=0$$ Integrating factor $\mu(xy)=\frac 1 {xy}$ $$2x\ln(y)+(x^2+1)\frac {y'}{y}=0$$ $$(x^2+1)'\ln(y)+(x^2+1)(\ln y)'=0$$ $$((x^2+1)(\ln y))'=0$$ Integrate: $$(x^2+1)\ln y=K$$

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For the integrating factor : split the problem so that: $$\mu (xy)=\mu(x)\mu(y)$$ Now we want to have: ( the derivative according to y:) $$(\mu (y) y \ln y)'=\mu (y)$$ And also here the derivative is taken according to x: $$2x^2\mu(x)=(\mu (x) (x^3+x))'$$ for this one I find that $$\frac {\mu(x)'}{\mu (x)} =- \frac 1 x $$ $$\implies \mu (x) =\frac 1x$$ For the other one I find that: $$(\mu (y) y \ln y)'=\mu (y)$$ $$\mu' (y) y \ln y+\mu (y)( \ln y + 1)=\mu (y)$$ $$\implies \frac {\mu '(y)}{\mu (y) } =- \frac 1 y \implies \mu (y)= \frac 1 y $$ Therefore: $$\mu (xy) = \mu (x) \mu (y) =\frac 1 {xy}$$

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You can also solve it the general way $$\frac {\partial}{\partial y}(M2x^2y \ln y)=\frac {\partial}{\partial x}( M(x^3+x))$$ We obtain here the derivative is taken according to $xy$: $$2x^2(xy M' \ln y +M(\ln (y) +1))=M'y(x^3+x)+M(3x^2+1)$$ Rearranging terms $$M(2x^2 \ln (y)-x^2-1)= M'(x^3y+xy-2x^3y \ln (y))$$ $$M(2x^2 \ln (y)-x^2-1)= M'xy(x^2+1-2x^2\ln (y))$$ It can be simplified as: $$M= -M'xy \implies (\ln (M))'=-\frac 1 {xy}$$ $$\implies M(xy)=\frac 1 {xy}$$

Answer 2

Considering $$m=m(xy)$$ we have $m_x=ym'$ and $m_y=xm'$

The expression that they have is correct.

The question is that how does the expression help in finding the integrating factor?

Answer 3

This is a separable equation so integrating factors aren't necessary; we have $y'(x) = -\frac{2xy(x)\log(y(x))}{x^2+1}$, and dividing both sides by $y(x)\log(y(x))$ yields $$ \frac{y'(x)}{\log(y(x))y(x)} = -\frac{2x}{x^2+1}.\tag 1 $$ Integrating both sides of $(1)$ with respect to $x$, we have $$\int \frac{y'(x)}{\log(y(x))y(x)}\ \mathsf dx = \int -\frac{2x}{x^2+1}\ \mathsf dx,\tag 2$$ and computing the integrals we see that $(2)$ is equivalent to $$ \log\log y(x) = -\log(x^2+1) + C, $$ where $C$ is an arbitrary constant. Hence $$ y(x) = e^{C/(x^2+1)}. $$

Answer 4

You are considering the equation $$2x^2y\ln(y)+(x^3+x)y'=0$$ Make life simpler and, to get rid of the unpleasant $\ln(y)$, let $y=e^z$ which reduces the equation to $$(1+x^2)z'+2x z=0\implies z=\frac C {1+x^2}\implies y=\exp\left(\frac C {1+x^2} \right)$$

Answer 5

and arrive at

$$m′⋅(2x^3y\ln(y)−yx^3−yx)=m⋅(3x^2+1−2x^2\ln(y)−2x^2)$$

But perhaps this is wrong anyways.

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No it's not wrong at all.

Note that $3x^2-2x^2=x^2$ $$m′⋅(2x^3y\ln(y)−yx^3−yx)=m⋅(3x^2+1−2x^2\ln(y)−2x^2)$$ $$m′⋅(2x^3y\ln(y)−yx^3−yx)=m⋅(x^2+1−2x^2\ln(y))$$ Then factorize $xy$ $$m′xy(2x^2\ln(y)−x^2−1)=m⋅(x^2+1−2x^2\ln(y))$$ Simplify $$m′xy=-m$$ $$\frac {m'}{m}=-\frac 1 {xy}$$ Integrate $$\int \frac {dm}{m}=-\int \frac {dxy} {xy}$$ $$m=\frac {C} {xy}$$ Choose $C=1$ : $$m=\frac {1} {xy}$$