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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

4 - x^{2} (\frac{d y}{d x}) = y^{2}, y = 5, x = 1

x d x + (y e^{- x}) d y = 0

x y^{^{'}} + 8 y + 7 = 7 x^{2} y^{^{'}} + 7 y

y^{^{'}} = sin (y) \cdot tan (y)

\frac{d y}{d x} = (\frac{x + 1}{y})

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