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

The solution for (1+2y)dx-(4-x)dy=0 is y=(c_{1})/(2(16-8x+x^2))-1/2

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

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

y = \frac{c _{1}}{2 ( 16 - 8 x + x ^{2} )} - \frac{1}{2}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{2 ( 16 - 8 x + x ^{2} )} - \frac{1}{2}

y = \frac{c _{1}}{2 ( 16 - 8 x + x ^{2} )} - \frac{1}{2}

\frac{\partial}{\partial x} (x d y)

d er i v a t i v e - e^{x} + 3 x - 3 sin (x)

\frac{d}{d x} ((sin (x^{2}) + 1)^{\frac{1}{y}})

\frac{\partial}{\partial x} (sin (π (6 x - 3 y)))

\frac{\partial}{\partial x} (x y - 5 x^{2} y)

What is the solution for (1+2y)dx-(4-x)dy=0 ?
The solution for (1+2y)dx-(4-x)dy=0 is y=(c_{1})/(2(16-8x+x^2))-1/2