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

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

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

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

y = \frac{1}{2} - \frac{1}{x} + \frac{c _{1}}{x ^{2}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{2} - \frac{1}{x} + \frac{c _{1}}{x ^{2}}

y = \frac{1}{2} - \frac{1}{x} + \frac{c _{1}}{x ^{2}}

\int 3 cos (x) - \frac{1}{x ^{3}} d x

\frac{d}{d x} (m (x - s) e^{k (x - d)} + p (x, d))

x \to \infty lim (3 x^{2} + 2)

\frac{d}{d x} (3 (e^{x} cos (x) - e^{x} sin (x)))

x \to 2 + lim (\frac{∣ x - 2 ∣}{x})

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