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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\int \frac{5 x - 1}{( x + 1 ) ( x - 2 )} d x

x^{9}, at x = 1

\int \frac{4}{( x - 1 ) ( x + 3 )} d x

f (x) = x^{2} x^{3} + 1, at x = 5

\frac{4 s}{( s ^{2} + 16 )}

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