Solve Function Problem: Steps Explained & Book Recommendation

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In summary, the problem is solved in steps by taking the derivative with respect to x and substituting y for x.
  • #1
asdf1
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there's an example question in a book I'm studying
"please solve
(2*x to the 3rd power +y to the 3rd power)dx-3x*y to the 2nd power*dy=0"
i have the steps to the solution but i don't know why the problem is solved in thosed steps. Can someone explain?
p.s.
can someone recommend a good book for math for someone in a physics major?
 
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  • #2
What exactly are you meant to be solving for? Show the steps you have..
 
  • #3
(2x3+ y3)dx- 3xy2dy= 0

This is a first order differential equation. It's almost exact since (2x3+ y3)y= 3y2 while (-3xy2)x= -3y2, differing only in the sign. Such an equation always has an "integrating factor"- a function such that, if you multiply the entire equation by the function the equation becomes exact but there is no general way of finding that integrating factor.

Here, because the terms involve only powers of x and y, I decided to try an integrating factor of the form xnym. Then the test for exactness becomes (2x3+nym+ xny3+m)y= 2mx3+nym-1+(3+m)xny2+m= (-3x1+ny2+m)x= -3(1+n)xny2+m.

That is, we must have 2mx3+nym-1+(3+m)xny2+m= -3(1+n)xny2+m.
The second term on the left has the same powers as the right but we need to get rid of the first term- OK, take m= 0! That "kills" the first term and to make the terms left equal we need to have 3+m= -3(1+n) or 3= -3(1+n) since m=0. That gives
1+n= -1 or n= -2.

Let's try that: multiplying (2x3+ y3)dx- 3xy2dy= 0
by x-2 gives (2x+ x-2y3)dx- 3x-1y2dy= 0.
(2x+ x-2y3)y= 3x-2y2 and
(-3x-1y2)x= 3x-2y2.
Yes! Those are the same so this equation is exact.

That MEANS that there exist some function, F(x,y), such that dF= (2x+ x-2y3)dx- 3x-1y2dy or, same thing, that
Fx= 2x+ x-2y3and Fy= - 3x-1y2 (you might remember that from multi-variable calculus).

Since Fx= 2x+ x-2y3, taking the anti-derivative (with respect to x), F(x,y)= x2- x-1y3+ g(y).
(The partial derivative with respect to x treats y like a constant. Taking the "anti-derivative" we treat y like a constant so that "constant of integration" may be any function of y.)

From that, Fy= -3x-1y2+ g'(y) and that must be equal to - 3x-1y2 . Comparing the two, that tells us that g'(y)= 0 or that g really is a constant.

IF F(x,y)= x2- x-1y3+ C, for any constant C, then
dF=(2x+ x-2y3)dx- 3x-1y2dy = 0. Since dF= 0, F is actually a constant: F(x,y)= x2- x-1y3+ C= C' where C' is also a constant, not necessarily equal to C. Since subtracting C from C' will still be an unknown constant, we can just write that as "C" and have
x2- x-1y3= C as the general solution to the differential equation.
 
  • #4
Another line of reasoning is:
(2x3+ y3)dx- 3xy2dy= 0

looking at the differential equation we see that it is
homogeneous (F(tx, ty) = tnF(x,y)). In this case,
each of the terms has degree 3.
So let's do a change of variables y = vx.

(2x3 + (vx)3)dx - 3x(vx)2(vdx + xdv) = 0
which gives us

x3( (2+v3) - 3v2(v + xv')) = 0
2 + v3 - 3v3 - 3xv2v' = 0
2(1 - v3) - x 3v2 v' = 0

which is seperable:
2 (dx/x) = 3v2 dv/(1 - v3)

2 d(ln(x)) + d( ln(1-v3)) = 0
2 ln(x) + ln(1-v3) = C'

ln( x2(1-v3) ) = C'
x2(1-v3) = C

switching back to y:
x2(1- (y/x)3) = C
x2- x-1y3 = C

as before.
 
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  • #5
How do you know to make the change of variables y = vx?
 
  • #6
qbert said:
looking at the differential equation we see that it is
homogeneous (F(tx, ty) = tnF(x,y)). In this case,
each of the terms has degree 3.
So let's do a change of variables y = vx.

That's a standard substitution for homogeneous equations. The point is that in a homogeneous equation we can always rewrite the equation so x and y only appear as [tex]\frac{y}{x}[/tex].

I missed that- qbert's solution is much simpler than mine!
 
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  • #7
but why would someone think of to make that substitution?
 

1. What is a function problem?

A function problem is a mathematical problem that involves finding the input value(s) that would result in a specific output value. It is essentially finding the missing piece of a mathematical equation.

2. What are the steps to solving a function problem?

The steps to solving a function problem can vary depending on the specific problem, but generally they involve identifying the given information, determining the type of function, setting up the equation, solving for the missing variable, and checking the solution to ensure it is valid.

3. Can you explain the steps to solving a function problem in more detail?

Sure! After identifying the given information and determining the type of function, you will set up an equation using the function notation and substitute in the given values. Then, you will solve for the missing variable by isolating it on one side of the equation. Finally, you will check your solution by plugging it back into the original equation and ensuring it makes sense in the context of the problem.

4. Are there any tips for solving function problems more efficiently?

Yes, one tip is to always double check your work and make sure you have correctly set up the equation and substituted in the values. Another tip is to practice solving different types of function problems to become more familiar with the process and recognize patterns.

5. Do you have a book recommendation for learning how to solve function problems?

A great book for learning how to solve function problems is "College Algebra Demystified" by Rhonda Huettenmueller. It provides clear explanations, step-by-step examples, and practice problems to help you master function problems and other algebraic concepts.

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