How can I find the integrating factor for this non-exact differential equation?

In summary: Similarly, the derivative, with respect to y, of f(x2+ y2) is 2y f ' (x2+ y2). Since we want M_x= M_y, we set (1- y)f+ (x- xy)(2x)f'=(x^2+ y^2)+(y+ x^2)(2y)f' or (1- y)f= (x^2+ y^2)- (x- xy)(2x)f'. Now you can solve for f.
  • #1
kingwinner
1,270
0
1) Solve the initial value problem (x - xy) + (y + x2) dy/dx = 0, y(0)=2
Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x2+y2)


M(x,y) + N(x,y) y' = 0

Omitting the hint, I found out that
(Nx - My) / M = 3 / (1-y) which is just a function of y

So u(y) = exp∫ 3 / (1-y) dy = 1 / |1-y|3 must be an integrating factor which can make the given ODE exact.

But how can I cope with the absolute value? I can't integrate until I can get rid of it. Now, if I just take one of u(y) = 1 / (1-y) or u(y) = 1 / (y-1), is that OK? Will I be getting the same general solution?

Another thing is that the initial condition is x=0, y=2>1, so must we take u(y) = 1 / (y-1) and not u(y) = 1 / (1-y)?



Secondly, I am very interested in what integrating factor the hint is referring to, what does it mean and how can I find it?



Can someone help? I would really appreciate!:smile:
 
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  • #2
Can somebody please help me?
 
  • #3
kingwinner said:
1) Solve the initial value problem (x - xy) + (y + x2) dy/dx = 0, y(0)=2
Hint: The ODE is not exact, but can be made exact. There exists an integrating factor of the form u=u(x2+y2)


M(x,y) + N(x,y) y' = 0

Omitting the hint, I found out that
(Nx - My) / M = 3 / (1-y) which is just a function of y

So u(y) = exp∫ 3 / (1-y) dy = 1 / |1-y|3 must be an integrating factor which can make the given ODE exact.

But how can I cope with the absolute value? I can't integrate until I can get rid of it. Now, if I just take one of u(y) = 1 / (1-y) or u(y) = 1 / (y-1), is that OK? Will I be getting the same general solution?
Did you try it and see?


Another thing is that the initial condition is x=0, y=2>1, so must we take u(y) = 1 / (y-1) and not u(y) = 1 / (1-y)?
Again, why not try it? Certainly one would suspect, since you would up with a "1- y" in the denominator, that solutions will not cross the line y= 1. You might check to see if that is true. If so, then, since y(0)> 1 you could assume that 1-y is always negative and |1- y|= y- 1.



Secondly, I am very interested in what integrating factor the hint is referring to, what does it mean and how can I find it?
For the third time try- don't be afraid to just plug things in and see what happens. If v(x,y) is an "integrating factor", then multplying by it makes the equation exact. If you think that the integrating factor is a function of [itex]x^2+ y^2[/itex], try multiplying by [itex]f(x^2+ y^2)[/itex] and see what happens.

[itex](x - xy)f(x^2+ y^2) + (y + x^2)f(x^2+ y^2) dy/dx = 0[/itex]
Now [itex]M= (x- xy)f(x^2+ y^2)[/itex] so [itex]M_x= (1- y)f + (x- xy)(2x)f'[/itex] and [itex]M_y= (x^2+ y^2)+(y+ x^2)(2y)f'[/itex]. Set those equal and solve for f.


Can someone help? I would really appreciate!:smile:
 
  • #4
HallsofIvy said:
Did you try it and see?
Yes, I got the same answer no matter I take |1-y|=1-y or |1-y|=y-1. In general, will they always be the same? (If so, then I don't have to waste time checking both cases every time)



Again, why not try it? Certainly one would suspect, since you would up with a "1- y" in the denominator, that solutions will not cross the line y= 1. You might check to see if that is true. If so, then, since y(0)> 1 you could assume that 1-y is always negative and |1- y|= y- 1.
OK, so for this entire question, we should take |1- y|= y - 1



For the third time try- don't be afraid to just plug things in and see what happens. If v(x,y) is an "integrating factor", then multplying by it makes the equation exact. If you think that the integrating factor is a function of [itex]x^2+ y^2[/itex], try multiplying by [itex]f(x^2+ y^2)[/itex] and see what happens.

[itex](x - xy)f(x^2+ y^2) + (y + x^2)f(x^2+ y^2) dy/dx = 0[/itex]
Now [itex]M= (x- xy)f(x^2+ y^2)[/itex] so [itex]M_x= (1- y)f + (x- xy)(2x)f'[/itex] and [itex]M_y= (x^2+ y^2)+(y+ x^2)(2y)f'[/itex]. Set those equal and solve for f.
You mean Ny, right?
f is a function of two variables, x and y, so this should give rise to a "partial" differential equation, right? But when you write only f on the last line, it seems like just one variable, is this a valid step?
So in order to find the integrating factor u(x2+y2), I need to solve another ODE, right?
 
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  • #5
No, you do not get a partial differential equation. f is a function of a single variable, say f(u), and we have replaced u by x2+ y2.
The derivate, with respect to x, of f(x2+ y2), is 2x f ' (x2+ y2 where f ' is the ordinary derivative of f. I have used the chain rule: df/dx= (df/du)(du/dx).
 

What are exact differential equations?

Exact differential equations are a type of differential equation where the derivative of a function can be expressed as a linear combination of the variables and the function itself. This means that the solution to an exact differential equation can be found by using an integrating factor.

How do you know if a differential equation is exact?

A differential equation is considered exact if the partial derivatives of the equation's terms with respect to each variable are equal. In other words, if the equation can be written in the form M(x,y)dx + N(x,y)dy = 0, then it is exact if and only if ∂M/∂y = ∂N/∂x.

What is the process for solving an exact differential equation?

The process for solving an exact differential equation involves finding an integrating factor that will transform the equation into an exact form. This integrating factor is a function that is dependent on one of the variables in the equation. Once the equation is in exact form, it can be solved by integrating both sides and then solving for the constant of integration.

What are some real-world applications of exact differential equations?

Exact differential equations have many applications in physics, engineering, and finance. For example, they can be used to model the flow of fluids, the behavior of electrical circuits, and the growth of populations. They are also used in economics to model supply and demand and in chemistry to determine reaction rates.

What are some common methods for finding the integrating factor in an exact differential equation?

Some common methods for finding the integrating factor in an exact differential equation include the method of integrating factors, the method of direct integration, and the method of separation of variables. Other methods, such as the method of variations of parameters, can also be used in certain cases.

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