Homework Help

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

user profile pic

wiky1011 | Student, College Freshman | eNotes Newbie

Posted November 12, 2011 at 12:26 PM via web

dislike 0 like

Solve the initial value problem.

4 dy/dx - 2x = xy^5   , y(1) = 3

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 26, 2011 at 4:04 PM (Answer #1)

dislike 0 like

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`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes