Differential eqn that has a uknown function of x

[ankur29]

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))

 

[Pengwuino]

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?

 

[ankur29]

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

 

[HallsofIvy]

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:

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

 

[ankur29]

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:

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

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?