Solve the initial value problem.

4 dy/dx - 2x = xy^5 , y(1) = 3

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You must integrate both sides to find y and x:

`int 4dy/(y^5+2) = int xdx `

`y^5 + 2 = (y^(5/2))^2 + (sqrt 2)^2`

`y^5 + 2 = (sqrt 2)^2*[(y^(5/2))^2/(sqrt 2)^2 + 1]`

`int 4dy/(y^5+2) = int 4dy/(sqrt 2)^2*[(y^(5/2))^2/(sqrt 2)^2 + 1]`

`int 4dy/(y^5+2) = 2 arctan (y^(5/2)/sqrt 2) + c`

`Calculate int xdx= x^2/2 + c`

`2 arctan (y^(5/2)/sqrt 2) = x^2/2 + c`

`arctan (y^(5/2)/sqrt 2) = x^2/4 + c`

`` `y^(5/2)/sqrt 2 = tan (x^2/4) + c`

`` `y^(5/2) = sqrt 2*tan (x^2/4) + c`

`y = (sqrt 2*tan (x^2/4))^(2/5) + c`

**Answer: y =`(sqrt 2*tan (x^2/4))^(2/5) + c` **

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