What is the solution for (2xy+x^2+x^4)dx-(1+x^2)dy=0 ?

The solution for (2xy+x^2+x^4)dx-(1+x^2)dy=0 is y=x(1+x^2)-arctan(x)(1+x^2)+c_{1}(1+x^2)

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(2xy+x^2+x^4)dx-(1+x^2)dy=0

(2 x y + x^{2} + x^{4}) d x - (1 + x^{2}) d y = 0

y = x (1 + x^{2}) - arctan (x) (1 + x^{2}) + c_{1} (1 + x^{2})

Solution steps

(2 x y + x^{2} + x^{4}) d x - (1 + x^{2}) d y = 0

Solve linear ODE:

y = x (1 + x^{2}) - arctan (x) (1 + x^{2}) + c_{1} (1 + x^{2})

y = x (1 + x^{2}) - arctan (x) (1 + x^{2}) + c_{1} (1 + x^{2})

\frac{\partial}{\partial x} (x^{(\frac{y}{z})})

35.4 cos (0.1 (x + 6.2))

x \to 1.01 lim (\frac{1}{x ^{2} - 1})

\frac{d}{d x} (1 + 3 x + 9 x^{2})

f (x) = 5 x^{2}

What is the solution for (2xy+x^2+x^4)dx-(1+x^2)dy=0 ?
The solution for (2xy+x^2+x^4)dx-(1+x^2)dy=0 is y=x(1+x^2)-arctan(x)(1+x^2)+c_{1}(1+x^2)