Question

常微分题目求解

Answer 1

u = y' = dy/dx

y'' = u' = du/dx = (dy/dx) * (du/dy) = u*(du/dy)

2u^2 = (y-1)*u*(du/dy)

∫ dy/(y-1) = ∫ du/(2u)

ln|y-1| = (1/2)*ln|2u| + C

y-1 = A*√(2u) ........ A=e^C

(1/B) * (y-1)^2 = u = y' = dy/dx .......... B=2A^2

∫ dx = ∫ B*(y-1)^(-2) dy

x = D - B/(y-1)

y = 1 + B/(D-x)

x=1, y=2, y' = (1/B) * (y-1)^2 = 1/B = -1
B = -1

x=1, y=2,
2=1+B/(D-1)
D = B+1 = 0

y = 1+1/x

( y' = -x^(-2), y'' = 2x^(-3) )