What is the solution for x(2y-3)dx+(x^2+1)dy=0 ?

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

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x(2y-3)dx+(x^2+1)dy=0

x (2 y - 3) d x + (x^{2} + 1) d y = 0

y = \frac{c _{1}}{2 ( x ^{2} + 1 )} + \frac{3}{2}

Solution steps

x (2 y - 3) d x + (x^{2} + 1) d y = 0

Solve separable ODE:

y = \frac{c _{1}}{2 ( x ^{2} + 1 )} + \frac{3}{2}

y = \frac{c _{1}}{2 ( x ^{2} + 1 )} + \frac{3}{2}

x \to 1 lim (7 x + a)

y^{^{'}} = \frac{2 y}{x}, y (2) = 4

(2 y^{2} + 6 x) d x + 2 x y d y = 0

\frac{d}{d x} (ln (\frac{x + 6}{x - 6}))

\int \frac{tan ^{2} ( 7 x )}{csc ( 7 x )} d x

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