What is the solution for (1+x^4)dy+x(1+4y^2)dx=0 ?

The solution for (1+x^4)dy+x(1+4y^2)dx=0 is y=(tan(-arctan(x^2)+c_{1}))/2

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(1+x^4)dy+x(1+4y^2)dx=0

(1 + x^{4}) d y + x (1 + 4 y^{2}) d x = 0

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

Solution steps

(1 + x^{4}) d y + x (1 + 4 y^{2}) d x = 0

Solve separable ODE:

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

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

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What is the solution for (1+x^4)dy+x(1+4y^2)dx=0 ?
The solution for (1+x^4)dy+x(1+4y^2)dx=0 is y=(tan(-arctan(x^2)+c_{1}))/2