briyola

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These are replies submitted by briyola

please can you help me to solve . thnak you to every one.

Hello everyone, I'm sorry for the inconvenience , but I already got stuck and couldn't solve this equation so please help me.

first of all we have N(q)=(exp(q)/sqrt(q))*sqrt(3/2)*D(3*q/2) . (1)

with: D(3*q/2)=exp(-3*q/2)*integral(t^2)*dt with t varie from 0 to sqrt(3*q/2) this integral is known as Dawson's name (2)

then we have q=(24/(n-3))*(x*k*m)/t with : t=0.6,n=6,k=1.4, so q=18.66666667*x*m (3)

then we have m = 1/2*[(exp(q)/(q*N(q)))-1-1/q] (4)

we have : q varied from 0 to infinity (0..infinity) and m varied from 0 to 1 (0..1)

 we have too:

t:=0.6:k:=1.4, n0:=6: n2:=6: x0:=0.495

p(x,t)=x*((t/(1-x/x0))-(12*x/(n0-3))-((k*m*12*x)/(n2-3))) (5)

v(x,t)= t*((ln(x/(x0-x)))+1/(1-x/x0))-(24*x/(n0-3))-t*ln(N(q))  (6)

the system we want to solve is:

p(x,t)=p(y,t) (7)

v(x,t)=v(y,t)

 

 the quetion is : first we must find the expression of m (4) as a function of x only. then we must find the system (7) solution
that is, we have to find the value of x and y. ( so  here we  search the value of  x  and y when m  varied )

(we have to fixe t :=0.6 and we  have to varie m from 0 to 1 .
for each  value of m we have a precise value of x and y )   ( x=0..0.17, y=0.171..0.50)
Finally, thank you evrey one here.

Kitonum14458, thank you very very much , for your answer  and your effort .

Acer17617  I  really thank you very very much for your answer  and your effort .

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