# Differential Equation Help

#### Dao Tuat

1. The problem statement, all variables and given/known data
Solve the following differntial equation

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

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

3. 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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#### arildno

Homework Helper
Gold Member
Dearly Missed
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

Staff Emeritus
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.

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#### 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?

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