What is the solution for (1+2y)dx+(4-x^2)dy=0 ?

The solution for (1+2y)dx+(4-x^2)dy=0 is y=(e^{(-ln(\frac{x+2)/2)+ln((x-2)/2)-c_{1})/(2)}}{2}-1/2

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(1+2y)dx+(4-x^2)dy=0

(1 + 2 y) d x + (4 - x^{2}) d y = 0

y = \frac{e ^{\frac{- l n ( \frac{x + 2}{2} ) + l n ( \frac{x - 2}{2} ) - c _{1}}{2}}}{2} - \frac{1}{2}

Solution steps

(1 + 2 y) d x + (4 - x^{2}) d y = 0

Solve separable ODE:

y = \frac{e ^{\frac{- l n ( \frac{x + 2}{2} ) + l n ( \frac{x - 2}{2} ) - c _{1}}{2}}}{2} - \frac{1}{2}

y = \frac{e ^{\frac{- l n ( \frac{x + 2}{2} ) + l n ( \frac{x - 2}{2} ) - c _{1}}{2}}}{2} - \frac{1}{2}

What is the solution for (1+2y)dx+(4-x^2)dy=0 ?
The solution for (1+2y)dx+(4-x^2)dy=0 is y=(e^{(-ln(\frac{x+2)/2)+ln((x-2)/2)-c_{1})/(2)}}{2}-1/2