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

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

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

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

y = \frac{e ^{2 c_{1}} x ^{2}}{2} + \frac{e ^{2 c_{1}}}{2} - \frac{1}{2}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{2 c_{1}} x ^{2}}{2} + \frac{e ^{2 c_{1}}}{2} - \frac{1}{2}

y = \frac{e ^{2 c_{1}} x ^{2}}{2} + \frac{e ^{2 c_{1}}}{2} - \frac{1}{2}

(8 x - 1) y^{^{'}} + 8 y = 0

cos (y) d x + e^{- x} sin (y) d y = 0

\frac{d}{df} (H (f)) = \frac{1 + u ( f ) ^{2}}{1 + f ^{2}}

d y - 2 y d x = 4 d x

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

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