What is the solution for y^{''}=(y^')^2+1 ?

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

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y^{''}=(y^')^2+1

y^{^{''}} = (y^{^{'}})^{2} + 1

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

Solution steps

y^{''} (t) = (y^{'} (t))^{2} + 1

Solve using substitution:

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

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

\int + 2 t d t

y = x^{2} - 1, y = - x + 2

\int 2 x^{2} x d x

h \to 0 lim (\frac{1 + h}{h})

\int_{0}^{π} 1

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