What is the solution for (dy)/(dx)=(sqrt(y))/(2x+1) ?

The solution for (dy)/(dx)=(sqrt(y))/(2x+1) is y=(ln^2(2x+1)}{16}+\frac{c_{1}ln(2x+1))/4+(c_{1}^2)/4

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(dy)/(dx)=(sqrt(y))/(2x+1)

\frac{d y}{d x} = \frac{y}{2 x + 1}

y = \frac{ln ^{2} ( 2 x + 1 )}{16} + \frac{c _{1} ln ( 2 x + 1 )}{4} + \frac{c _{1}^{2}}{4}

Solution steps

\frac{d y}{d x} = \frac{y}{2 x + 1}

Solve separable ODE:

y = \frac{ln ^{2} ( 2 x + 1 )}{16} + \frac{c _{1} ln ( 2 x + 1 )}{4} + \frac{c _{1}^{2}}{4}

y = \frac{ln ^{2} ( 2 x + 1 )}{16} + \frac{c _{1} ln ( 2 x + 1 )}{4} + \frac{c _{1}^{2}}{4}

y^{^{'}} - a^{2} k y + b^{2} = 0

y^{^{'}} - 2 t y^{3} = 0

\frac{d y}{d t} = - k y

y^{^{'}} = e^{- x - y}

\frac{d y}{d t} = (2 + 3 t) y

What is the solution for (dy)/(dx)=(sqrt(y))/(2x+1) ?
The solution for (dy)/(dx)=(sqrt(y))/(2x+1) is y=(ln^2(2x+1)}{16}+\frac{c_{1}ln(2x+1))/4+(c_{1}^2)/4