Question

find solution pi of y''=-1/2y^(2) satisfying pi(0)=1,pi'(0)=-1

Answer 1

Is this (1) y"=-½y² or (2) y"=-1/(2y²)? Both possibilities will be considered.

Let z=dy/dx=y', y"=z'; (dz/dy)(dy/dx)=dz/dx; y"=z'=dz/dx=z(dz/dy).

(1) z(dz/dy)=-½y².

∫zdz=-½∫y²dy, z²/2=-⅙y³+a where a is a constant. 

z²=-⅓y³+a, dy/dx=√(a-⅓y³).

y'(0)=-1, so -1=√(a-⅓(y(0))³)=√(a-⅓), because y(0)=1, a-⅓=1, a=4/3.

∫dy/√(4/3-⅓y³)=∫dx. This may not be conventionally integrable.

(2) z(dz/dy)=-1/(2y²).

z²/2=-½∫dy/y², z²=-∫y^-2dy,

z²=1/y+a, y'=z=√(1/y+a) or y'=-√(1/y+a). y(0)=1, y'(0)=-1⇒a=0. Only the negative square root applies.

dy/dx=-y^-½, because a=0.

∫y^½dy=-∫dx, ⅔y^3/2=-x+b, so b=⅔, because y(0)=1, and ⅔y^3/2=⅔-x, or y^3/2=1-3x/2.

CHECK

y'√y=-1, y"√y+y'(y'/(2√y)=0, y"√y+(-1/√y)²(1/2√y)=0, y"+1/(2y²)=0, y"=-1/(2y²). Checks out.

It would appear that the solution to y"=-1/(2y)² is y^3/2=1-3x/2, or maybe y=∛(9x²/4-3x+1). However, these two equations are not exactly equivalent, though similar, as can be clearly seen if the equations are graphed. This is due to the fact that only the square roots of positive numbers are valid; y=∛(9x²/4-3x+1) contains two symmetrical halves, while y^3/2=1-3x/2 only has the left half of the same curve. For example, there is no real value for y when x>⅔. The unique solution is therefore y^3/2=1-3x/2.