求微分方程yy"+1=(y')²通解
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解：
令 y'= p
y*p*(dp/dy)- p^2 + 1 = 0
p*dp/(p^2 - 1) = dy/y
ln[/(p^2 - 1/] = ln[C*y^2]
ln[/(p^2 - 1/] = ln[C*y^2]
p^2 = C1*y^2 + 1
p = k* (C1*y^2 + 1 )^(1/2), ( k = ±1 )
dy/(C1*y^2 + 1 )^(1/2) = k*dx
1) C1 = 0 时
dy = k*dx
y = k*x + C
2) C1 = C^2 > 0 时, (C > 0 )
ln[ y + ( y^2 + 1/C^2 ) ] + C2 = k*C*x + C2
3) C1 = - C^2 < 0 时, (C > 0 )
arcsin(Cy) = k*C*x + C2
( k = ±1 )
令 y'= p
y*p*(dp/dy)- p^2 + 1 = 0
p*dp/(p^2 - 1) = dy/y
ln[/(p^2 - 1/] = ln[C*y^2]
ln[/(p^2 - 1/] = ln[C*y^2]
p^2 = C1*y^2 + 1
p = k* (C1*y^2 + 1 )^(1/2), ( k = ±1 )
dy/(C1*y^2 + 1 )^(1/2) = k*dx
1) C1 = 0 时
dy = k*dx
y = k*x + C
2) C1 = C^2 > 0 时, (C > 0 )
ln[ y + ( y^2 + 1/C^2 ) ] + C2 = k*C*x + C2
3) C1 = - C^2 < 0 时, (C > 0 )
arcsin(Cy) = k*C*x + C2
( k = ±1 )
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