What is the solution for x^{''}+(x^')^2=0 ?

The solution for x^{''}+(x^')^2=0 is x=ln(-t+c_{1})+c_{2}

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x^{''}+(x^')^2=0

x^{^{''}} + (x^{^{'}})^{2} = 0

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

Solution steps

x^{''} (t) + (x^{'} (t))^{2} = 0

Solve using substitution:

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

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

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What is the solution for x^{''}+(x^')^2=0 ?
The solution for x^{''}+(x^')^2=0 is x=ln(-t+c_{1})+c_{2}