Differential eqn that has a uknown function of x

In summary, you are trying to find the constant A(x) in a quadratic equation. You first need to find the equation's roots, π and 0.04x+33. When you do that, you find that A(x) = (∏(0.04x+33)^2). So your final equation becomes V(x) = (∏(0.04x+33)^2)17.
  • #1
ankur29
10
0
Hi guys
i was wondering how one should tackle a differential eqn with multiple functions of x in it like this:
i.e d/dx(R(x)^2*V(x)*(V0-V(x))
 
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  • #2
You'll need to give a little more information than that. What is what you have equal to? Is it really the derivative of that whole product?
 
  • #3
Pengwuino said:
You'll need to give a little more information than that. What is what you have equal to? Is it really the derivative of that whole product?

sorry my bad

d/dx ( R(x)^2* V(x) * (Vo-V(x) ) =0

it is to be solved wrt 'V(x)'
i have gotten this from a problem related to a wind turbine
V(x) must be found as it describes velcoity in the turbine's wake at a distance x from the 1st turbine,so to estimate the windspeed available to the next turbine behind the first

V0 is constant
 
  • #4
That just tells you that R(x)^2* V(x) * (Vo-V(x))= C, a constant. Of course, there are an infinite number of function pairs, (V, R), that satisfy that. To solve for V depending on R, Multiply the R^2V(V0- V)= R^2V0 V- R^2V^2 to write it as a quadratic equation,
R^2V^2- R^2V0V+ C= 0 and use the quadratic formula:

[tex]V(x)= \frac{R(x)V_0\pm\sqrt{R(x)^2V_0^2- 4C}}{2R(x)}[/tex]
 
  • #5
HallsofIvy said:
That just tells you that R(x)^2* V(x) * (Vo-V(x))= C, a constant. Of course, there are an infinite number of function pairs, (V, R), that satisfy that. To solve for V depending on R, Multiply the R^2V(V0- V)= R^2V0 V- R^2V^2 to write it as a quadratic equation,
R^2V^2- R^2V0V+ C= 0 and use the quadratic formula:

[tex]V(x)= \frac{R(x)V_0\pm\sqrt{R(x)^2V_0^2- 4C}}{2R(x)}[/tex]

Thanks for replying
Here is what i have attempted:

d/dx ( R(x)^2* V(x) * (Vo-V(x) ) =0
In reality there should be a π before the R(X)^2= ∏(R(x)^2

Thus it was ought to be d/dx (∏(R(x)^2)*V(x)* (Vo-V(x) ) =0
Where R(x)= αx+r1=0.04x+33
∏(R(x)^2) =A(x) ‘area’ = (∏(0.04x+33)^2)

So rewriting this as a quadratic as you have said
As (∏(R(x)^2)*V(x)* (Vo-V(x) ) =constant
= (∏(R(x)^2)*V(x)* Vo - (∏(R(x)^2)*V(x)^2 +c
In form ax^2+bx+c=0
& For simplification
∏(R(x)^2) =A(x) ‘area’ = (∏(0.04x+33)^2)
0.04x+33
-A(x)V(x)^2 +A(x)VoV(x) +c =0

[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21%20%20-A%28x%29V%28x%29%5E2%20%2BA%28x%29VoV%28x%29%20%2Bc%20%3D0.gif

[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21V%28x%29%3D%5Cfrac%7B-%28A%28x%29Vo%5Cpm%5Csqrt%7B%28A%28x%29Vo%29%5E2-4%28-A%28x%29c%29%7D%7D%7B2%28-A%28x%29%29%7D.gif

and knowing that : A(x) ‘area’ = (∏(0.04x+33)^2)

[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21V%28x%29%3D%5Cfrac%7B-%28%28%5Cpi%280.04x%2B33%29%5E2%29Vo%5Cpm%5Csqrt%7B%28%28%5Cpi%280.04x%2B33%29%5E2%29Vo%29%5E2-4%28-%28%5Cpi%280.04x%2B33%29%5E2%29c%29%7D%7D%7B2%28-%28%5Cpi%280.04x%2B33%29%5E2%29%29%7D.gif


How do I find the constant, I know that with boundary considtions x=0 V(x)=17
When I try I get C=0 and if I use that in my expression for V(x) I get 17 always at varying x


can you see what i am doing wrong?
 
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1. What is a differential equation with an unknown function of x?

A differential equation with an unknown function of x is an equation that involves the derivative of an unknown function with respect to x. The solution to this type of equation is a function that satisfies the equation for all values of x.

2. Why is an unknown function of x used in differential equations?

Unknown functions of x are used in differential equations because they are a powerful tool for modeling and solving problems in many scientific fields. By using an unknown function, we can express the relationship between quantities that are continuously changing.

3. What is the process for solving a differential equation with an unknown function of x?

The process for solving a differential equation with an unknown function of x involves finding the general solution, which is a family of functions that satisfies the equation. This is done by integrating both sides of the equation and then solving for the arbitrary constant.

4. What are some real-world applications of differential equations with an unknown function of x?

Differential equations with an unknown function of x are widely used in fields such as physics, chemistry, engineering, economics, and biology to model and predict the behavior of systems. Some examples include the growth of populations, the spread of diseases, and the motion of objects in space.

5. Are there any techniques for solving differential equations with an unknown function of x?

Yes, there are several techniques for solving differential equations with an unknown function of x, such as separation of variables, substitution, and the method of undetermined coefficients. These techniques can be used depending on the form of the equation and the initial conditions given.

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