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

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

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

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

y = tan (arctan (x) - c_{1})

Solution steps

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

Solve separable ODE:

y = tan (arctan (x) - c_{1})

y = tan (arctan (x) - c_{1})

\frac{d y}{d x} = 15 x sec (y)

y^{^{'}} = \frac{x ^{2}}{y + 1}, y (0) = 4

\frac{d x}{d t} = \frac{1}{x e ^{t + 11 x}}

y^{^{'}} = 2 y e^{t}

10 - x^{2} y^{^{'}} = e^{6 y} sin (x)

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