What is the solution to this separable differential equation?

In summary: This is a "separable" differential equation. There must exist a function F(x,y) such that \frac{\partial F}{\partial x}= y- 2x- 1 and \frac{\partial F}{\partial y}= x+ y- 4. Can you find F?
  • #1
Dao Tuat
16
0

Homework Statement


Solve the following differntial equation


Homework Equations


(x+y-4)dy - (2x-y+1)dy = 0

EDIT: Oops, it should be (x+y-4)dy - (2x-y+1)dx = 0

The Attempt at a Solution


I integrated both sides and ended up with:
[y(2x +y-8)]/2 = X^2 + x(1-y) + c

Did I solve this right, or am I supposed to do something else?
 
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  • #2
First of all: Which "dy" should be a "dx" instead??

Secondly, you have probably done this incorrectly by fallaciously regarding "x" and "y" as constants in the integration on both sides. This is wrong.
 
  • #3
Did you mean [tex] \left( {x + y - 4} \right)dy - \left( {2x - y + 1} \right)dx = 0 [/tex]? If so then you have neglected some products.

On the left, [tex] \int {xdy} \ne xy [/tex] and similarly on the right.
 
  • #4
If it is, in fact, (x+y-4)dy - (2x-y+1)dy = 0
which is the same as (x+y-4)dy + (-2x+y-1)dy = 0, then it
is an "exact" equation: (x+y-4)x= 1= (-2x+y-1)y.
There exist a function F(x,y) such that dF= (x+y-4)dy - (2x-y+1)dy = 0 and so F(x,y) is a constant. Do you know how to find that F?
 
  • #5
Aero said:
On the left, [tex] \int {xdy} \ne xy [/tex] and similarly on the right.

Isn't it? It's off by a constant, but that's it unless I'm crazy.
 
  • #6
d_leet said:
Isn't it? It's off by a constant, but that's it unless I'm crazy.

No, because x is not a constant. Consider d(xy). Because x is not a constant this is not just xdy.
 
  • #7
Thanks for everyones help. I'm still a bit confused though. Any tips would be very much appreciated.
 
  • #8
Aero said:
No, because x is not a constant. Consider d(xy). Because x is not a constant this is not just xdy.

You are integrating with respect to y, however, so as it is written above that will be equal to xy+c, since when you integrate with respect to y you would treat x as a constant.
 
  • #9
d_leet said:
You are integrating with respect to y, however, so as it is written above that will be equal to xy+c, since when you integrate with respect to y you would treat x as a constant.

But when you differentiate with respect to y, you get

d/dy (xy) = x + y * dx/dy

and x is not treated as a constant, so why should it be treated as a constant in integration?
 
  • #10
Aero said:
But when you differentiate with respect to y, you get

d/dy (xy) = x + y * dx/dy

and x is not treated as a constant, so why should it be treated as a constant in integration?

If you had differentiated correctly there would be no problem. You differentiated with respect to x even though you wrote it as you were going to differentiate with respect to y. If you differentiate xy with respect to y, you treat x as a constant because you are not differentiating with respect to it, hence the derivative of xy with respect to y is simply x.
 
  • #11
You cannot simply treat x as a constant in the term (x+y-4)dy and integrate it wrt y, since it is part of the differential equation. An equation can only be solved like that if it is separable into the form f(x)dx = g(y)dy, in which case both sides can be integrated. However, this equation cannot be separated into that form.

On how to solve this equation... I'm stumped!
 
  • #12
as aero suggested, consider d(xy)=xdy+ydx
 
  • #13
d_leet said:
If you had differentiated correctly there would be no problem. You differentiated with respect to x even though you wrote it as you were going to differentiate with respect to y. If you differentiate xy with respect to y, you treat x as a constant because you are not differentiating with respect to it, hence the derivative of xy with respect to y is simply x.

This is only true for partial differentiation. However, here we are dealing with normal differentiation and integration.

By your logic, d/dy (x) = 0 because you would treat x like a constant. But this is plain wrong: d/dy (x) = dx/dy.
 
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  • #14
We have had 4 people pointing out that d_leet was wrong. Dao Tuat, have you looked at my response. Can you answer my question?

This is a "separable" differential equation. There must exist a function F(x,y) such that [itex]\frac{\partial F}{\partial x}= y- 2x- 1[/itex] and [itex]\frac{\partial F}{\partial y}= x+ y- 4[/itex]. Can you find F?
 
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1. What is a differential equation?

A differential equation is a mathematical equation that involves an unknown function and its derivatives. It is used to describe the relationship between a quantity and its rate of change.

2. Why are differential equations important?

Differential equations are important because they are used to model and solve many real-world problems in fields such as physics, engineering, economics, and biology. They are also the foundation of many mathematical and scientific theories.

3. What are the different types of differential equations?

The main types of differential equations are ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs involve a single independent variable, while PDEs involve multiple variables.

4. How do you solve a differential equation?

There are various methods for solving differential equations, including separation of variables, substitution, and using integrating factors. It is important to first identify the type of differential equation and choose an appropriate method for solving it.

5. What are some real-world applications of differential equations?

Differential equations are used in many fields to model and solve problems, such as predicting population growth, analyzing the motion of objects, and understanding the behavior of electrical circuits. They are also used in the development of scientific theories, such as the theory of relativity and quantum mechanics.

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