Solve the given differential equation subject to the indicated initial condition. `(1+x^4)dy + x(1+4y^2)dx = 0 , y(1) = 0`

Solve the given differential equation subject to the indicated initial condition. 

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

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Solution:

To solve the differential equation `(1+x^4)dy + x(1+4y^2)dx = 0` with the initial condition `y(1) = 0`, we can first use separation of variables to get:

`∫ (1+x^4) dy = -∫ x(1+4y^2) dx`

Integrating both sides with respect to their respective variables gives:

`y(1+x^4) = -(1/4)x^2(1+4y^2) + C`

Solving for y gives:

`y = -(1/4)(x^2)(1+4y^2) + C*(1+x^4)^(-1)`

Now we can use the initial condition to find the value of C:

`y(1) = -(1/4)(1^2)(1+4y^2) + C*(1+1^4)^(-1) = 0`

`C = 1`

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So the general solution is:

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

And the solution to the differential equation subject to the initial condition is:

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