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}

f (x) = 13 x

\frac{d}{d x} (arcsin (7 x + 1))

\int (8 x^{3}) d x

\int 4 x cos (2 x) d x

x \to 2 lim (2 x + 4)

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