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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

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