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Thread: 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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    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 $\displaystyle \frac 12y^2 + 3y = \sin 5x + c$ to $\displaystyle y = \cdots $ use the quadratic formula ($\displaystyle a = \frac 12$, $\displaystyle b = 3$ and $\displaystyle c = \sin 5x + C$)
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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?
    $\displaystyle (3 + 2y) \, dy = 5\cos(5x) \, dx$

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

    $\displaystyle y(0) = 1$ ... $\displaystyle C = 4$

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

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

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

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

    since $\displaystyle y(0) = 1 $

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