What is the solution for y^'=(x(1+y^2))/(4y) ?

The solution for y^'=(x(1+y^2))/(4y) is y=sqrt(e^{(x^2)/4+c_{1)}-1},y=-sqrt(e^{(x^2)/4+c_{1)}-1}

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y^'=(x(1+y^2))/(4y)

y^{^{'}} = \frac{x ( 1 + y ^{2} )}{4 y}

y = e^{\frac{x ^{2}}{4} + c_{1}} - 1, y = - e^{\frac{x ^{2}}{4} + c_{1}} - 1

Solution steps

y^{'} (x) = \frac{x ( 1 + y ^{2} )}{4 y}

Solve separable ODE:

y = e^{\frac{x ^{2}}{4} + c_{1}} - 1, y = - e^{\frac{x ^{2}}{4} + c_{1}} - 1

y = e^{\frac{x ^{2}}{4} + c_{1}} - 1, y = - e^{\frac{x ^{2}}{4} + c_{1}} - 1

\frac{d y}{d x} y = (ln (x))^{l n (x)}

y^{^{'}} = \frac{5 x + 8}{y}

\frac{d y}{d x} = 2 x 36 - y^{2}

\frac{d y}{d x} = \frac{1}{7} (y)^{\frac{1}{2}} (cos (y^{\frac{1}{2}}))^{2}

y^{^{'}} = \frac{y ^{4}}{x ^{6}}, y (1) = 3

What is the solution for y^'=(x(1+y^2))/(4y) ?
The solution for y^'=(x(1+y^2))/(4y) is y=sqrt(e^{(x^2)/4+c_{1)}-1},y=-sqrt(e^{(x^2)/4+c_{1)}-1}