What is the solution for y^'= 1/(y^1) ?

The solution for y^'= 1/(y^1) is y=sqrt(2t+c_{1)},y=-sqrt(2t+c_{1)}

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y^'= 1/(y^1)

y^{^{'}} = \frac{1}{y ^{1}}

y = 2 t + c_{1}, y = - 2 t + c_{1}

Solution steps

y^{'} (t) = \frac{1}{y ^{1}}

Solve separable ODE:

y = 2 t + c_{1}, y = - 2 t + c_{1}

y = 2 t + c_{1}, y = - 2 t + c_{1}

\frac{d y}{d x} = 6 (5 - y)

\frac{d y}{d x} = \frac{2 y ( 2 x - 1 ) e ^{x}}{2 x + 1}

y^{^{'}} = 1 + 4 y^{2}, y (0) = 0

y^{^{'}} = \frac{( e ^{- x} - e ^{x} )}{( 3 + 4 y )}

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

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