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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

7 cos (7 x), 7 - 7 cos (7 x), [0, \frac{π}{7}]

5 y x^{2} + 1 \frac{d y}{d x} + 4 x = 0

\frac{d}{d x} (\frac{1}{( 3 x + 1 ) ^{2}})

y = ln (\frac{x ^{2}}{4})

\frac{d}{d x} (\frac{2 x ^{3} - 2}{3 x + 4})

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