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

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

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

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

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}

\frac{d x}{d t} = 12 \cdot x \cdot t

\frac{d y}{d x} = 12 y x^{2}

y = x^{3} \cdot e^{- 2 x} cos (3 x) \frac{d y}{d x}

(3 y) d x + (2 x) d y = 0

t^{- 1} \frac{d y}{d t} = 8 cos^{2} (y), y (2) = \frac{3 π}{4}

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