Question

Determination of Integrating Factors, need help here.

I'm stuck somewhere at partial derivative of M - partial derivative of N. I can't go on further because they don't show up the same as the given.
I might be doing something wrong.. Need help. :<

1. y(4+y)dx - 2(x^2-y)dy = 0
2. 3(x^2+y^2)dx + x(x^2+3y^2+6y)dy = 0
( ^ I'm almost at the answer, I just need to check
your answer too.)
3. 2(x-y+2)dx + x(x-2y+2)dy = 0
4. y(2x+y-2)dx - 2(x+y)dy = 0

Answer 1

$$y(4+y)dx-2(x^2-y)dy=0$$to check if it is exact consider:$$\frac{\partial}{\partial y}y(4+y)=4+2y\\\frac{\partial}{\partial x}2(x^2-y)=4x$$clearly it is not an exact equation yet; however, we may determine an integration factor \(\mu\) to turn this into one:$$\frac{\partial}{\partial y}\mu y(4+y)=(4+2y)\mu+y(4+y)\frac{\partial\mu}{\partial y}\\\frac{\partial}{\partial x}2\mu(x^2-y)=4x\mu+2(x^2-y)\frac{\partial\mu}{\partial x}$$since we want our equation exact, equate these two:$$(4+2y)\mu+y(4+y)\frac{\partial\mu}{\partial y}=4x\mu+2(x^2-y)\frac{\partial\mu}{\partial x}\\(4+2y-4x)\mu+y(4+y)\frac{\partial\mu}{\partial y}=2(x^2-y)\frac{\partial\mu}{\partial x}$$in general it is difficult to determine \(\mu\) as a function of both variables; instead, consider either the case where \(\mu\) is a function of only \(x\) i.e. \(\partial\mu/\partial y=0\) and so \(\partial\mu/\partial x=d\mu/dx\):$$(4+2y-4x)\mu=2(x^2-y)\frac{d\mu}{dx}\\\frac{d\mu}{dx}=\frac{4+2y-4x}{2x^2-2y}\mu$$or the case where \(\mu\) is a function of only \(y\) i.e. \(\partial\mu/\partial x=0\) and so \(\partial\mu/\partial y=d\mu/dy\) giving:$$(4+2y-4x)\mu+y(4+y)\frac{d\mu}{dy}=0\\\frac{d\mu}{dy}=-\frac{4+2y-4x}{4y+y^2}\mu$$

Answer 2

note that these two cases are special because they reduce to *ordinary* differential equations... :-)

Answer 3

I figured out that the 1, 3, and 4 is not applicable with the topic. Is it really legal to use partial derivative of u over partial derivative of x or y? Finding another way to solve those 3 is hard. （ ´,_ゝ`)

Answer 4

of course it's legal! those equations fall out from multiplying by our unknown integration factor \(\mu\) and assuming exactness... hmm, let's see though

Answer 5

can you try 3 and 4 for me? I just want to see more of it. Our exams might include this kind of rare solution from you. :D

Answer 6

Seriously, I didn't even know that partial derivative of u..

Answer 7

sorry for the delay, I was doing other problems

Answer 8

let's see...

Answer 9

$$2(x-y+2)dx + x(x-2y+2)dy = 0$$we want some \(u(x,y)\) that makes this exact:$$2u(x-y+2)dx+ux(x-2y+2)dy=0$$i.e.$$\frac{\partial}{\partial y}2u(x-y+2)=\frac{\partial}{\partial x}ux(x-2y+2)$$

Answer 10

let's try it out:$$\frac{\partial}{\partial y}2u(x-y+2)=2u(-1)+2(x-y+2)\frac{\partial u}{\partial y}\\\frac{\partial}{\partial x}u(x^2-2xy+2x)=u(2x-2y+2)+(x^2-2xy+2x)\frac{\partial u}{\partial x}\\-2u+2(x-y+2)\frac{\partial u}{\partial y}=u(2x-2y+2)+(x^2-2xy+2x)\frac{\partial u}{\partial x}\\u(2x-2y+4)=-(x^2-2xy+2x)\frac{\partial u}{\partial x}+2(x-y+2)\frac{\partial u}{\partial y}$$by inspection, you should recognize that since \(2x-2y+4=2(x-y+2)\) it is sufficient for \(\partial u/\partial x=0\) i.e. have \(u\) merely be a function of \(y\), making \(\partial u/\partial y=du/dy\) and yielding a comparatively simple ordinary differential equation in \(u\):$$u(2x-2y+4)=2(x-y+2)\frac{du}{dy}\\u=\frac{du}{dy}$$... which we know yields \(u=e^y\)

Answer 11

do you even assumed a value for u? o.o

Answer 12

no; the whole point is that we know what we want \(u\) to do and then solve a differential equation for it. in practice this boilds down to:$$u=\exp\left(\int\frac{\partial M/\partial y-\partial N/\partial x}Ndx\right),\exp\left(-\int\frac{\partial M/\partial y-\partial N/\partial x}Mdy\right)$$

Answer 13

we thus have that our left-hand side (after multiplying by \(e^y\)) is the exact differential of some potential function \(F\), i.e.:$$\frac{\partial F}{\partial x}=2e^y(x-y+2)=2xe^y-2ye^y+4e^y\\\frac{\partial F}{\partial y}=xe^y(x-2y+2)=x^2e^y-2xye^y+2xe^y$$

Answer 14

to solve for \(F\) we then just integrate:$$F(x,y)=\int\frac{\partial F}{\partial x}dx+g(y)=2\int(xe^y-ye^y+2e^y)~dx=x^2e^y-2xye^y+4xe^y+g(y)\\F(x,y)=\int\frac{\partial F}{\partial y}dy+f(x)=x\int (xe^y-2ye^y+2e^y)dy=xe^y-xy^2e^y+2xe^y+f(x)$$

Answer 15

oops that second should be \(x^2e^y-2xye^y+4xe^y+f(x)\)... apparently I can't integrate by parts correctly

Answer 16

anyways that gives an example potential function solution:$$F(x,y)=x^2 e^y-2xy e^y+4xe^y$$

Answer 17

so I just have to integrate that F(x,y) from the 2nd equationand the du/dy from the first equation?

sorry for being forceful, but I just about to sleep.

Answer 18

I'm just**

Answer 19

eh? you solve the differential equation of \(u\) to get an integration factor (that's what those \(\exp\) formulae essentially do -- solve those equations in one-step)

Answer 20

then you multiply by \(u\) to get a usual exact differential equation and then integrate \(\int M dx,\int Ndy\) to determine the potential function

Answer 21

the idea is that \(dF=Mdx+Ndy\) hence \(Mdx+Ndy=0\) becomes \(dF=0\) which in turn gives \(F(x,y)=C\) i.e. level curves of our potential

Answer 22

I mean, this dμ/dy=−4+2y−4x4y+y2μ
and F(x,y)=x2ey−2xyey+4xey... Should I just integrate it? Are we're just getting the I.F. after integrating it? I'm getting kind of confused.

Answer 23

the integration factor comes from solving:$$u(2x-2y+4)=2(x-y+2)\frac{du}{dy}$$, which, you should note, is just:$$\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)u=M\frac{du}{dy}\\\frac{\partial N/\partial x-\partial M/\partial y}{M}u=\frac{du}{dy}\\\frac1udu=\frac{\partial N/\partial x-\partial M/\partial y}{M}dy\\\log u=\int\frac{\partial N/\partial x-\partial M/\partial y}{M} dy\\u=\exp\left(\int\frac{\partial N/\partial x-\partial M/\partial y}{M} dy\right)$$

Answer 24

then we multiply our equation by \(u\) to make it exact, and solve by integrating both \(M,N\) to determine a potential function

Answer 25

oh cool. thanks. I really have to sleep now. I'll just try to do what I can.. :<