# Homework Help: Differential equation (I really need help)

1. Jul 2, 2005

### Beez

Q1: Find the general solution for [x^5*y^7 - x^3*y]dx + [x^6*y^6 + x^4]dy = 0
I tried to find IF by doing the following but I got stuck since I cannot obtain a solution with one valiable. What should I do or how can I obtain IF with different method ( I tried to do y = vx, v + x*dv/dx = M/N, too but did not work either) ?

d(x^5*y^7 - x^3*y)/dy = 7*x^5*y^6 - x^3
d(x^6*y^6 + x^4)/dx = 6*x^5*y^6 + 4*x^3

[(6*x^5*y^6 + 4*x^3) - (7*x^5*y^6 - x^3)]/(x^5*y^7 - x^3*y)
= [1*(5 - x^2*y^6)]/[y*(x^2*y^6 - 1)] --> still has both x and y

[(7*x^5*y^6 - x^3)- (6*x^5*y^6 + 4*x^3)]/(x^6*y^6 + x^4)
= [1*(x^2*y^6 -5)]/[x*(x^2*y^6 + 1)] --> still has both x and y

I'm taking a correspondence course, so this is the only place I can get help. So someone please help me!

2. Jul 2, 2005

### Corneo

I believe the equation is seperable.

3. Jul 2, 2005

### Beez

Could you tell me where to start?

How can I separate $$\frac{1(5-x^2y^6)}{y(x^2y^6 - 1)}$$?
I tried the following but it did not even help a bit.

$$\frac{1(5-x^2y^6)}{y(x^2y^6 - 1)}$$
= $$\frac{5}{y(x^2y^6 - 1)}$$ -$$\frac{x^2y^6}{y(x^2y^6- 1)}$$
=$$\frac{5}{y(x^2y^6 - 1)}$$ -$$\frac{x^2y^5}{(x^2y^6 - 1)}$$
= $$\frac{x^2y^7 - y}{5}$$ - $$\frac{x^2y^6 - 1}{x^2y^5}$$
=$$\frac{x^2y^7}{5}$$ - $$\frac{y}{5}$$ - $$\frac{x^2y^6 }{x^2y^5}$$ - $$\frac{1}{x^2y^5}$$
=$$\frac{x^2y^7}{5}$$- $$\frac{y}{5}$$-y - $$\frac{1}{x^2y^5}$$ $$\rightarrow$$ does not lead me anywhere...

Would you please give me a hint how I could start separating them?

Last edited: Jul 2, 2005
4. Jul 2, 2005

### James R

This is a homogeneous type differential equation. Put:

$$y = u(x) x$$

so that

$$\frac{dy}{dx} = u + x \frac{du}{dx}$$

The D.E. is:

$$[x^{12} u^7 - x^4 u] dx = - [x^{12} u^6 + x^4] dy$$

So

$$\frac{dy}{dx} = -\frac{x^{12} u^7 - x^4 u}{x^{12} u^6 + x^4} = u + x \frac{du}{dx}$$,

which, after a little algebra, leads to...

$$-2 x^{11} u^7 = \frac{du}{dx}$$

Integrating gives:

$$-\frac{1}{6} x^{12} + c = \frac{1}{8} \left(\frac{y}{x}\right)^8$$

$$-\frac{1}{6} x^{20} + c x^8 = \frac{1}{8} y^8$$

Final solution:

$$y = \left(-\frac{4}{3} x^{20} + k x^8\right)^{1/8}$$

Last edited: Jul 3, 2005
5. Jul 2, 2005

### Beez

Sorry but I need a further explanation...

James, thank you for your help but few points are still not clear to me.
Hope you don't mind my asking you to help me out again.
$$\frac{dy}{dx} = -\frac{x^12u^7 - x^4u}{x^12u^6+x^4}=u + x\frac{du}{dx}$$ (sorry I don't know how to superscript the first digit of two digits exponent, so please read $$x^12$$ for x^12 and $$x^11$$ for x^11 and so on.)
become
$$-2x^11u^7$$ = $$\frac{du}{dx}$$ after a little algebra; but instead I got the following:

$$\frac{dy}{dx} = -\frac{x^12u^7 - x^4u}{x^12u^6+x^4}=u + x\frac{du}{dx}$$
-->
$$\frac{x^4u-x^12u^7 -x^12u^7 -x^4u}{x^12u^6+x^4}= x\frac{du}{dx}$$

$$\frac{-2x^12u^7}{x^12u^6+x^4} * \frac{1}{x} = \frac{du}{dx}$$

$$\frac{-2x^11u^7}{x^12u^6+x^4} = \frac{du}{dx}$$

Why couldn't I get rid of the denominator like you did?

In addition, the general solution that was given in the studying material was
$$x^4y^6 - 2x^2 = Cy^2$$
I tried to obtain the above solution from your solution, but could not do it...

Last edited: Jul 2, 2005
6. Jul 2, 2005

### mathelord

ACCORDING TO WHAT I READ BEFORE,IF YOU HAVE A DE LIKE PDX+QDY=O,
THE INTEGRATING FACTOR IS THE EXPONENTIAL OF THE INTEGRAL OF (1/Q)[dp/dy -dq/dx]. you still need to confirm this,because i am 16 and might not give you a correct information.

7. Jul 2, 2005

### saltydog

Writing the equation as:

$$\frac{dy}{dx}=-\frac{(x^5y^7-x^3y)}{x^6y^6+x^4}$$

I do not get equality when I back-substitute your solution. Also, your solution does not appear to agree with numerical calculations for the case y(0.1)=1.

8. Jul 2, 2005

### Beez

Mathload, yes you are right. I was suppose to be able to get IF by using the formula you provided. I also tried $$\frac{1}{P}[\frac{dq}{dx}-\frac{dp}{dy}]$$, but did not work either. Please take a look at the following and offer me suggestions where I should go from there.

Original equation: $$[x^5y^7-x^3y]dx + [x^6y^6+x^4]dy = 0$$

$$\frac{1}{x^6y^6+x^4}(7x^5y^6-x^3-6x^5y^6-4x^3) = \frac{x^5y^6-5x^3}{x^6y^6 + x^4} = \frac{x^3(x^2y^6-5)}{x^4(x^2y^6+1)}=\frac{1(x^2y^6-3)}{x(x^2y^6+1)}$$
* I could not reduce to one variable form.

$$\frac{1}{x^5y^7-x^3y}(6x^5y^6+4x^3-7x^5y^6+x^3) = \frac{-x^5y^6+5x^3}{x^5y^7-x^3y} = \frac{x^3(5-x^2y^6)}{x^3y(x^2y^6-1)}=\frac{5-x^2y^6}{(x^2y^6-1)}$$
* I could not reduce to one variable form for this one either.

Thanks.

9. Jul 3, 2005

### James R

Looks like I made a couple of mistakes. Hmm...

10. Jul 3, 2005

### lurflurf

Thats fun!
First group terms of the same order together
x^5y^6(ydx+xdy)+x^3(-ydx+xdy)=0
now we see the first term should be in xy and the second in y/x so
we see this because we know
d(xy)=ydx+xdy
x^2 d(y/x)=-ydx+xdy so
x^5y^6d(xy)+x^5d(y/x)=0
so we need to multiply by x^n y^m for some n&m so that each term is an exact differential
x^(5+n)y^(6+m) d(xy)+x^(5+n) y^m d(y/x)=0
we need
5+n=6+m
-m=5+n
so n=-2 m=-3 so
x^3y^3 d(xy)+(y/x)^-3 d(y/x)=0
Integrate
x^4y^4/4-(y/x)^-2/2=C

now do these for extra practice
(x^3y^2+xy^4-x^2y^3)dx+(x^4y+x^2y^3+x^3y^2)dy=0
and
ydx+(x+x^3y^2)dy=0

Last edited: Jul 3, 2005
11. Jul 3, 2005

### saltydog

What, fun watching us squirm cus' we couldn't solve it? (just funnin')
I'm satisfied. It works out. Thanks a bunch for showing that.

yep, I did.

Beez, you got that? Can you post all the solutions? Plots too (just use any initial conditions. I used y(0.1)=1)? Just a suggestion.

Last edited: Jul 3, 2005
12. Jul 3, 2005

### mathelord

now since it's exact i mean mdx+ndy,u the main function should equal the integral of m wrt x.Now u will have a constant,which u will get by differentiating u with repect to y and equating it with n.there u have a variable for ur consatnt.give it a try.

13. Jul 3, 2005

### lurflurf

If you can write an exact differential in the form f(u(x,y))d[u(x,y)] as I did above you can save some work by integrating directly. You only need to integrate with respect to one varivable then differentiate and match when the exact difererential is not in that form.

14. Jul 3, 2005

### Beez

Sorry that I'm not that smart, but...

Lurflurf, thanks for showing me how to solve the problem.
I understood most of them, but could not understand $$x^2d(\frac{y}{x})=-ydx+xdy$$ part in your

Last edited: Jul 3, 2005
15. Jul 3, 2005

### James R

$$df(x,y) = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$$

so, if $f(x,y) = y/x$ we have:

$$df = -\frac{y}{x^2}dx + \frac{1}{x}dy$$

and therefore

$$x^2 df = -ydx + x dy$$

16. Jul 3, 2005

### lurflurf

That is too bad that is not in your textbook. I hope you can see it is true and that you mean how would I think of it. In solving these types of problems knowing a few differentials helps to solve the problems efficiently, but as others mentioned there is something you could do if a differential you do not recognize appears. You hope UMdx+UNdy is exact for some multiplier U. Here M=-y N=x so -yUdx+xUdy is we hope exact if would be a pretty general method if we assume U(x,y) but we do not have the tools to deal with this case so assume (Dy)U=0 the dervatives in what follows are partial. Thus if we integrate xU we get xUy+V. V is the integrating constant and it may depend on x. Now to make sure -yUdx+xUdy is an exact differential we assume the function it is the differential of has equal mixed partials that is (Dx)(Dy)=(Dy)(Dx) holds for this function. So -yU=(Dx)(xUy+V)=Uy+xU'y+V'=(U+xU')y+V'. So V'=0 -U=U+xU' this is seperatable with U=1/x^2 any (particular solution will do). So we knew d(xUy)=-yUdx+xUdy now that we know U=1/x^2 we know d(y/x)=-(y/x^2)dx+(1/x)dx or clearing denominators
x^2d(y/x)=-ydx+xdx as desired. You can see that knowing this fact makes the problem easier, but it is not strictly necsisary. I think the key point of this problem is sometimes an inexact differential can be written as the sum of some exact differentials.

Last edited: Jul 3, 2005
17. Jul 4, 2005

### mathelord

lurluf,thats exactly what i suggested,but in another form

18. Jul 4, 2005

### lurflurf

I know you did, but you confused me a bit. At first you tried to treat the whole equation as exact (or possible a multiple of exact). Then later after I solved it you were talking about solving the equation in a confusing way. I will try to outline what I think confused people in this problem. First off is the method you want to use cannot be used directly. Every first order equation has an integrating factor, but it will not always be easy to find. If we assume an integrating factor u(x,y) it is found by solving a partial differential equation that if often harder than the ordinary differential equation we are trying to solve. Thus we assume u(x) or u(y) or some special form if we have reason to belive it would work. For this problem it is easy to see (having solved it) that x^-2y^-3 is an integrating factor. This I.F. depends on x and y so we can not find it the way you suggest. We could guess u=x^ny^m, which is what I did a few step later, but it is not clear to me that it would be known at the very begining. Thus I wrote the equation as a sum of known differentials multiplied by functions.
that is
x^5y^6(ydx+xdy)+x^3(-ydx+xdy)=0
Then I could see x^-2y^-3 would be an I.F. This method is finding an integrating factor by inspection, because we can see what it should be.
To do this we needed to know that
d(xy)=ydx+xdy
and
x^2 d(y/x)=-ydx+xdy
If we did not know this we could find out as I did in the previous post using the method you discribe.
However if we do know these differentials we do not need to use that method as we chose the I.F. by inspection so that the differentials are exact. This puts the equation in the form
f(u)du+g(v)dv
In this form it can be integrated. Also we have used an integrating factor and do not need one anymore.
Remember you can find the I.F. by integrating differentiating and matching unless an integrating factor exist in the form u(x) or u(y). In the cases where u(x,y) is needed we could
1. Solve a PDE (bad ideal in general)
2. Try a substitution (fair ideal if you are clever)
3. Try to use inspection (good ideal if you spot known differentials)
4. Try to guess the form of the I.F. (can be tricky)

I hope this clears it up.

19. Jul 4, 2005

### saltydog

Hello Lurflurf,

I'm interested in that statement. In the DE forum below someone asks how to find the integrating factor for:

$$pydx +qxdy +x^my^n(rydx+sxdy)=0$$

If we assume an integrating factor f(x,y), what is the PDE associated with finding the integrating factor?

20. Jul 4, 2005

### lurflurf

That is an interesting equation.
I noticed xyd(log(x^py^q))=pydx+qxdy
but that does not help much. Perhaps a substitution is in order? Anyway for the PDE I will consider something more general.
Mdx+Ndy
let u be an integrating factor then
$$\frac{\partial}{{\partial}y}(uM)=\frac{\partial}{{\partial}x}(uN)$$
$$u \frac{{\partial}M}{{\partial}y}+M \frac{{\partial}u}{{\partial}y}=u \frac{{\partial}N}{{\partial}x}+N \frac{{\partial}u}{{\partial}x}$$
or
$$u (\frac{{\partial}M}{{\partial}y}-\frac{{\partial}N}{{\partial}x}})=N \frac{{\partial}u}{{\partial}x}-M \frac{{\partial}u}{{\partial}y}$$
As you can see this is no better than the ODE and is perhaps worse.

21. Jul 4, 2005

### saltydog

Jesus Lurflurf, I hate it when people have to "remind me" (cus' I should know already) something that's on the first few pages of a math book (an ODE text in this case). Thanks a bunch for taking the time for inputting all the LaTex. Great though, so we can then define the integrating factor for the equation I asked you about. Here, I'll put it in LaTex for punish work for not remembering the stuff up there. For the equation:

$$pydx +qxdy +x^my^n(rydx+sxdy)=0$$

The integrating factor u(x,y) will satisfy the following first order PDE with variable coefficients:

$$(qx+svx^my^n)\frac{\partial u}{\partial x}-(py+rx^my^{m-1})\frac{\partial u}{\partial y}-\left[x^{m-1}y^m(p-q)(xr-svm)\right]u=0$$

assuming p,q,r,s,v,m,n are constants. Great. Thus, the original question I asked in the DE forum is how do you prove a first order ODE in the form:

$$\frac{dy}{dx}=-\frac{M(x,y)}{N(x,y)}$$

has an integrating factor u(x,y), then reduces to showing that a first-order PDE of the following form:

$$N(x,y)\frac{\partial u}{\partial x}-M(x,y)\frac{\partial u}{\partial y}-\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)u=0$$

has a solution. Well, why didn't they say that over there?