What is the solution for y^{''}=(y^')^2 ?

The solution for y^{''}=(y^')^2 is y=-ln(t+c_{1})+c_{2}

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y^{''}=(y^')^2

y^{^{''}} = (y^{^{'}})^{2}

y = - ln (t + c_{1}) + c_{2}

Solution steps

y^{''} (t) = (y^{'} (t))^{2}

Solve using substitution:

y = - ln (t + c_{1}) + c_{2}

y = - ln (t + c_{1}) + c_{2}

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\frac{d}{d x} (\frac{x - 8}{x ^{2} - 4})

What is the solution for y^{''}=(y^')^2 ?
The solution for y^{''}=(y^')^2 is y=-ln(t+c_{1})+c_{2}