x=ln(t+√(1+t^2) (1)
∫(1->y) e^(u^2) du +∫(t->0) sinu/√(1+u^2) du =0 (2)
To find: d^2y/dx^2|t=0
solution:
∫(1->y) e^(u^2) du +∫(t->0) sinu/√(1+u^2) du =0
两边求导对t
dy/dt. e^(y^2) - sint/√(1+t^2) =0
dy/dt =[ sint/√(1+t^2)].e^(-y^2)
x=ln(t+√(1+t^2)
两边求导对t
dx/dt
=[1/(t+√(1+t^2)] . [ 1+ t/√(1+t^2)]
=1/√(1+t^2)
dy/dx
=(dy/dt)/(dx/dt)
=sint.e^(-y^2)
d/dt(dy/dx)
=(cost -2ysint. dy/dt ).e^(-y^2)
=(cost -2ysint. 【[ sint/√(1+t^2)].e^(-y^2) 】 ).e^(-y^2)
=[ cost -2y(sint)^2.e^(-y^2) ] .e^(-y^2)/√(1+t^2)
d^2y/dx^2
=d/dt(dy/dx) / (dx/dt)
=[ cost -2y(sint)^2.e^(-y^2) ] .e^(-y^2)
from (2)
∫(1->y) e^(u^2) du +∫(t->0) sinu/√(1+u^2) du =0
t=0, y=1
from (1)
x=ln(t+√(1+t^2)
t=0, x=0
t=0 , (x,y)=(0,1)
d^2y/dx^2 |t=0
=[ cos0 -0 ] .e^(-1)
=1/e