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

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

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(dy)/((1+4y^2))=-x/((1+x^4))dx

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

y = \frac{tan ( - arctan ( x ^{2} ) + c _{1} )}{2}

Solution steps

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

Solve separable ODE:

y = \frac{tan ( - arctan ( x ^{2} ) + c _{1} )}{2}

y = \frac{tan ( - arctan ( x ^{2} ) + c _{1} )}{2}

6 x^{\frac{3}{2}}

x \to \infty lim (0.9 9^{x})

\frac{d}{d t} (a t e^{t})

3 x y^{2} y^{^{'}} + 3 y^{3} = 1

\frac{d}{d x} (- 2 e^{- x})

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