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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

x \to 0 lim (\frac{1}{x} \cdot sin (x))

y^{^{'}} = x y

\frac{d}{d x} (7 x^{2} - 12 x)

\frac{d}{d x} (\frac{3}{2} x - 5)

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