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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

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