eq:=(1/6)*7^(1/2)*6^(1/2)*hypergeom([1/14], [15/14], 1/(1/y0)^(7/3))/(y0^(1/6)*(1/y0)^(1/6))-hypergeom([1/14], [15/14], 1)-1

fsolve(eq, y0 = 1 .. 2)

1.7417148480262249059

answer.mw

curve := [2*t*(3*t^4+50*t^2-33)/(t^2+1)^3, (2*(7*t^6-60*t^4+15*t^2+2))/(t^2+1)^3]

eq := [x = op(1, curve), y = op(2, curve)]

sort(op([2, 1], eliminate(eq, t)), [x, y], ascending)

 For R>0 and a>0.

((a^(1/2)*ln(1-a+2*(1-R)^(1/2)*(-a-R)^(1/2)-2*R)-I*ln(-(2*I)*a^(1/2)*(1-R)^(1/2)*(-a-R)^(1/2)+(a-1)*R-2*a)+((1/2)*I)*ln(a))*(-1+R)^(1/2)*(-a-R)^(1/2)+(a+R)^(1/2)*(I*ln((2*I)*a^(1/2)*(1+R)^(1/2)*(-a+R)^(1/2)+(a-1)*R+2*a)-a^(1/2)*ln(1-a+2*R+2*(1+R)^(1/2)*(-a+R)^(1/2))-((1/2)*I)*ln(a))*(1-R)^(1/2))/((a+R)^(1/2)*(1-R)^(1/2)*a^(1/2))

Copy code and paste to Maple.It should work.

integral.mw

Mariusz Iwaniuk

Hi

You can use a "prime" from palletes. :)

2D_is_ok..mw

M_Iwaniuk

Use a MultiSeries package.

I_Mariusz

Read this :http://www.mapleprimes.com/questions/200812-Error-in-Dsolvenumericbvp-Initial#

Only_5.mw

I_Mariusz

Numerically this integral can be calculated.


See atached file.Numeric.mw

I_Mariusz

f(2)=-pi/2 is a correct answer.See atached file.

Maybe You want f(x)=abs(arcsin(1-x)), then f(2)=Pi/2  ,f>0  ?

Answer.mw

I_Mariusz

Yours homework.

PDE.mw