What is the solution for 2y+3+(2x-4)(dy)/(dx)=0 ?

The solution for 2y+3+(2x-4)(dy)/(dx)=0 is y=(c_{1})/(2(2x-4))-3/2

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2y+3+(2x-4)(dy)/(dx)=0

2 y + 3 + (2 x - 4) \frac{d y}{d x} = 0

y = \frac{c _{1}}{2 ( 2 x - 4 )} - \frac{3}{2}

Solution steps

2 y + 3 + (2 x - 4) \frac{d y}{d x} = 0

Solve separable ODE:

y = \frac{c _{1}}{2 ( 2 x - 4 )} - \frac{3}{2}

y = \frac{c _{1}}{2 ( 2 x - 4 )} - \frac{3}{2}

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

3 x y^{^{'}} - 2 y = 0, y (0) = x^{\frac{2}{3}}

cot (x) (e - y + 1) d x = (1 + sin^{2} (x)) d y

y^{^{'}} arctan (x) - y (1 + x^{2})^{- 1} = 0

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

What is the solution for 2y+3+(2x-4)(dy)/(dx)=0 ?
The solution for 2y+3+(2x-4)(dy)/(dx)=0 is y=(c_{1})/(2(2x-4))-3/2