O Omegachi Diamond Member Mar 27, 2001 3,922 0 76 Sep 29, 2005 #1 i am totally stuck on this problem and don't know where to start: verify that both phi(t)=1-t and psi(t)=-(t^2)/4 are solutions of the IVP y'= (-t + sqrt(t^2+4y))/2 , y(2) = -1 if anyone knows what to do, please help me
i am totally stuck on this problem and don't know where to start: verify that both phi(t)=1-t and psi(t)=-(t^2)/4 are solutions of the IVP y'= (-t + sqrt(t^2+4y))/2 , y(2) = -1 if anyone knows what to do, please help me
S Suspicious-Teach8788 Lifer Feb 19, 2001 20,155 23 81 Sep 29, 2005 #3 Sorry buddy. There's a reason why I got a D+ in that course and I'm taking it again... Except we're only on linear algebra so far..
Sorry buddy. There's a reason why I got a D+ in that course and I'm taking it again... Except we're only on linear algebra so far..
Random Variable Lifer Aug 10, 2001 10,420 2 0 Sep 29, 2005 #4 let y = 1 - t then y' = -1 and (-t + sqrt(t^2 + 4 - 4t))/2 = (-t + sqrt((t - 2)^2))/2 = (-t + t - 2)/2 = -1 and obviously y(2)=-1
let y = 1 - t then y' = -1 and (-t + sqrt(t^2 + 4 - 4t))/2 = (-t + sqrt((t - 2)^2))/2 = (-t + t - 2)/2 = -1 and obviously y(2)=-1
B BRObedoza Diamond Member Apr 16, 2004 4,133 0 76 Sep 29, 2005 #5 just take the first derivative and substitute into the diff. eq.