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Math Help - solving for y

  1. #1
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    solving for y

    dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1

    I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lord12 View Post
    dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1

    I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?
    as i am feeling lazy today, i'll assume your solution is correct so far. as far as going from \frac 12y^2 + 3y = \sin 5x + c to y = \cdots use the quadratic formula ( a = \frac 12, b = 3 and c = \sin 5x + C)
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  3. #3
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    Quote Originally Posted by lord12 View Post
    dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1

    I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?
    (3 + 2y) \, dy = 5\cos(5x) \, dx

    3y + y^2 = \sin(5x) + C

    y(0) = 1 ... C = 4

    3y + y^2 = \sin(5x) + 4

    y^2 + 3y + \frac{9}{4} = \sin(5x) + 4 + \frac{9}{4}

    \left(y + \frac{3}{2}\right)^2 = \sin(5x) + \frac{25}{4}

    y + \frac{3}{2} = \pm \sqrt{\sin(5x) + \frac{25}{4}}

    since y(0) = 1

    y = \sqrt{\sin(5x) + \frac{25}{4}} - \frac{3}{2}
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