What is the solution for (dy)/(dx)=(y^2)/(2x+3) ?

The solution for (dy)/(dx)=(y^2)/(2x+3) is y=-2/(ln(3+2x)+c_{1)}

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(dy)/(dx)=(y^2)/(2x+3)

\frac{d y}{d x} = \frac{y ^{2}}{2 x + 3}

y = - \frac{2}{ln ( 3 + 2 x ) + c _{1}}

Solution steps

\frac{d y}{d x} = \frac{y ^{2}}{2 x + 3}

Solve separable ODE:

y = - \frac{2}{ln ( 3 + 2 x ) + c _{1}}

y = - \frac{2}{ln ( 3 + 2 x ) + c _{1}}

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\int \frac{sec ^{2} ( x )}{3 + tan ^{2} ( x )} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = t, y (0) = 1, y^{^{'}} (0) = 0

\int_{0}^{1} \frac{x - 8}{x ^{2} - 5 x + 6} d x

\frac{d}{d x} (\frac{x ^{2}}{ln ( x )})

What is the solution for (dy)/(dx)=(y^2)/(2x+3) ?
The solution for (dy)/(dx)=(y^2)/(2x+3) is y=-2/(ln(3+2x)+c_{1)}