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Question: how do i solve the boundary conditions

Question:how do i solve the boundary conditions

umar khan 30 Maple
differential-equation numeric bvp
December 01 2016
0


 

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 (1)
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G := 6.6743*10^(-8); 1; c := 2.99792458*10^10; 1; pi := 3.143; 1; rho := 5.3808*10^14

0.6674300000e-7

  

0.2997924580e11

  

3.143

  

0.5380800000e15

 (2)
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diff(P(r), r) = -G*(rho*c^2+P(r))*((4*pi*r^3*(1/3))*rho+4*Pi*r^3*P(r)/c^2)/(c^2*(r^2-2*G*r*(4*pi*r^3*(1/3))*rho/c^2)), diff(v(r), r) = 1.485232054*10^(-28)*((4*pi*r^3*(1/3))*rho+4.450600224*10^(-21)*Pi*r^3*P(r))/(r^2-1.485232054*10^(-28)*r*(4*pi*r^3*(1/3))*rho)

diff(P(r), r) = -0.7426160269e-28*(0.4836021866e36+P(r))*(0.2254913920e16*r^3+0.4450600224e-20*Pi*r^3*P(r))/(r^2-0.3349070432e-12*r^4), diff(v(r), r) = 0.1485232054e-27*(0.2254913920e16*r^3+0.4450600224e-20*Pi*r^3*P(r))/(r^2-0.3349070432e-12*r^4)

 (3)
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condition; -1; P(0) = 0, v(1014030) = -.4283

P(0) = 0, v(1014030) = -.4283

 (4)
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I found the solution of P(r) at P(0)=0, but could obtain the result of v(r) at v(1014030)=-0.4283, v(r) may have a graph such that i can goes from -0.4283 to 0.
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