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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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