What is the solution for (dy}{dx}=\frac{1+y^2)/x ?

The solution for (dy}{dx}=\frac{1+y^2)/x is y=tan(ln(x)+c_{1})

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(dy}{dx}=\frac{1+y^2)/x

\frac{d y}{d x} = \frac{1 + y ^{2}}{x}

y = tan (ln (x) + c_{1})

Solution steps

\frac{d y}{d x} = \frac{1 + y ^{2}}{x}

Solve separable ODE:

y = tan (ln (x) + c_{1})

y = tan (ln (x) + c_{1})

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

1 - x^{2} y^{^{'}} = 1 - y^{2}

y^{^{'}} = y (1 - \frac{y}{10})

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

y^{^{'}} sec (t) = \frac{- y ^{3}}{2}

What is the solution for (dy}{dx}=\frac{1+y^2)/x ?
The solution for (dy}{dx}=\frac{1+y^2)/x is y=tan(ln(x)+c_{1})