What is the solution to this separable differential equation?

[Dao Tuat]

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?

 

[arildno]

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.

 

[Aero]

Did you mean \left( {x + y - 4} \right)dy - \left( {2x - y + 1} \right)dx = 0? If so then you have neglected some products.

On the left, \int {xdy} \ne xy and similarly on the right.

 

[HallsofIvy]

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?

 

[d_leet]

On the left, \int {xdy} \ne xy and similarly on the right.

Isn't it? It's off by a constant, but that's it unless I'm crazy.

 

[Aero]

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.

 

[Dao Tuat]

Thanks for everyones help. I'm still a bit confused though. Any tips would be very much appreciated.

 

[d_leet]

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.

 

[Aero]

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?

 

[d_leet]

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.

 

[cristo]

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!

 

[tim_lou]

as aero suggested, consider d(xy)=xdy+ydx

 

[Aero]

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.

 

[HallsofIvy]

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 \frac{\partial F}{\partial x}= y- 2x- 1 and \frac{\partial F}{\partial y}= x+ y- 4. Can you find F?