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Let y=f(x)
f''(x) = √(f(x)
Multiplying both sides of above equation by f'(x) and then integrating One gets
(f'(x)^2)/2 = 2/3 *(f(x)^3/2)
By using initial conditions one finds that the constant of integration=0
We have f'(x) = (2/√3)f(x)^3/4
dy/dx = (2/√3)y^3/4
Thus y^-3/4 dy = (2/√3) dx
Integrating both sides one gets
4y^1/4 =( 2/√3)*x + c
It is given that y=1 at x=0.
By putting the relation in above equation we get c=4
4y^1/4 = (2/√3)x + 4
Put x=2√3 to get y=16