What is the solution for y^{''}+(y^')^2+1=0 ?

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

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y^{''}+(y^')^2+1=0

y^{^{''}} + (y^{^{'}})^{2} + 1 = 0

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

Solution steps

y^{''} (t) + (y^{'} (t))^{2} + 1 = 0

Solve using substitution:

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

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

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a re a \frac{1}{8} x, 8 x, \frac{2}{x}

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(x^{2} + y) d x - (x) d y = 0

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