What is the solution for f^'(x)=-1/(f(x)),f^{''}(x)=-f(x) ?

The solution for f^'(x)=-1/(f(x)),f^{''}(x)=-f(x) is f(x)=sqrt(-2x+c_{1)},f(x)=-sqrt(-2x+c_{1)}

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f^'(x)=-1/(f(x)),f^{''}(x)=-f(x)

f^{^{'}} (x) = - \frac{1}{f ( x )}, f^{^{''}} (x) = - f (x)

f (x) = - 2 x + c_{1}, f (x) = - - 2 x + c_{1}

Solution steps

f^{'} (x) = - \frac{1}{f ( x )}

Solve separable ODE:

f (x) = - 2 x + c_{1}, f (x) = - - 2 x + c_{1}

f (x) = - 2 x + c_{1}, f (x) = - - 2 x + c_{1}

\frac{d}{d x} ((y)^{2}) = y^{2} x

\frac{d y}{d x} = \frac{x ^{3} + 1}{y}, y = 2, x = 1

\frac{d y}{d x} y - 6 x^{3} + cos (x) = 0

y^{^{'}} = 3 y - 15

y^{^{'}} = \frac{x y ^{3}}{5}, y (0) = 5

What is the solution for f^'(x)=-1/(f(x)),f^{''}(x)=-f(x) ?
The solution for f^'(x)=-1/(f(x)),f^{''}(x)=-f(x) is f(x)=sqrt(-2x+c_{1)},f(x)=-sqrt(-2x+c_{1)}