Solve the initial value problem: 4*dy/dx - 2x = x*y^5, y(1)=3

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Add 2y to both sides:

`4dy/dx = x*y^5 + 2x`

`` `4dy/dx = x*(y^5 + 2)`

Multiply by `dx =gt 4dy = x*(y^5 + 2)*dx`

`` Divide by `y^5 + 2 =gt 4dy/(y^5 + 2) = x*dx`

Integrate to find y:

`int 4dy/(y^5 + 2) = intx*dx`

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

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

`int dy/(y^5 + 2) = int dy/{[y^(5/2)]^2 + (sqrt 2)^2}`

`` `int dy/(y^5 + 2) = (sqrt 2/2)*arctan ((y^(5/2))/sqrt2)`

`4 int dy/(y^5 + 2) = (2*sqrt 2)*arctan ((y^(5/2))/sqrt2)`

`4 int dy/(y^5 + 2) = x^2/2 + c =gt`

`=gt (2*sqrt 2)*arctan ((y^(5/2))/sqrt2) = x^2/2 + c` `arctan ((y^(5/2))/sqrt2) =sqrt2*x^2/8 + c`

`(y^(5/2))/sqrt2 = tan (sqrt2*x^2)/8 + c`

`y^(5/2) =sqrt 2*tan (sqrt2*x^2)/8 + c`

y = `[sqrt 2*tan (sqrt2*x^2)/8]^(2/5) + c`

y(1)=3 = > `c = 3 -[sqrt 2*tan (sqrt2)/8]^(2/5)`

**ANSWER: y = `[sqrt 2*tan (sqrt2*x^2)/8]^(2/5) ` + c**

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