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Math Help - Separation of variables

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    Separation of variables

    Solve the separable differential equation y'(x) = sqrt(2y(x)+15)
    and find the particular solution satisfying the initial condition y(1) = -3
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    Quote Originally Posted by tttcomrader View Post
    Solve the separable differential equation y'(x) = sqrt(2y(x)+15)
    and find the particular solution satisfying the initial condition y(1) = -3
    y' = \sqrt{2y+15}

    \frac{y'}{\sqrt{2y+15}} = 1

    \int \frac{y'dx}{\sqrt{2y+15}} = \int 1 dx

    Continue.
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    Thanks, I first thought the x is actually y times x instead of the function of y with variable x.

    Well, I got y = (x^2 - 7)/2

    Sounds right?
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    Quote Originally Posted by tttcomrader View Post
    Thanks, I first thought the x is actually y times x instead of the function of y with variable x.

    Well, I got y = (x^2 - 7)/2

    Sounds right?
    Is y^{\prime} = \sqrt{2y + 15}?

    Let me give you a hand:
    \int \frac{y}{\sqrt{2y + 15}}~dy

    Let u = 2y + 15 \implies du = 2 dy

    \int \frac{y}{\sqrt{2y + 15}}~dy = \int \frac{u - 15}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{dy}{2}

    = \frac{1}{4} \int \left ( \sqrt{u} - <br />
\frac{15}{\sqrt{u}} \right ) du

    Can you take it from here?

    -Dan
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    Yeah, looks like I made a silly mistake up there.

    Now, I get to a point where I found C = {2,-4}.

    So y = [(x+2)^2 - 15]/2 or y = [(x-4)^2 - 15]/2

    Are they both the right answers?
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    Quote Originally Posted by tttcomrader View Post
    Yeah, looks like I made a silly mistake up there.

    Now, I get to a point where I found C = {2,-4}.

    So y = [(x+2)^2 - 15]/2 or y = [(x-4)^2 - 15]/2

    Are they both the right answers?
    Quote Originally Posted by topsquark View Post
    Is y^{\prime} = \sqrt{2y + 15}?

    Let me give you a hand:
    \int \frac{y}{\sqrt{2y + 15}}~dy

    Let u = 2y + 15 \implies du = 2 dy

    \int \frac{y}{\sqrt{2y + 15}}~dy = \int \frac{u - 15}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{dy}{2}

    = \frac{1}{4} \int \left ( \sqrt{u} - <br />
\frac{15}{\sqrt{u}} \right ) du

    Can you take it from here?

    -Dan
    Let me continue.
    = \frac{1}{4} \int \left ( \sqrt{u} - <br />
\frac{15}{\sqrt{u}} \right ) du

    = \frac{1}{4} \left ( \frac{2}{3}u^{3/2} + 2 \cdot 15u^{1/2} \right ) + C

    = \frac{1}{6}(2y + 15)^{3/2} + \frac{15}{2}(2y + 15)^{1/2} + C

    So we have the differential equation:

    y^{\prime} = \sqrt{2y + 15}

    \frac{dy}{\sqrt{2y + 15}} = dx

    \int \frac{dy}{\sqrt{2y + 15}} = \int dx

    The LHS is just the integral I finished up at the top, so
    \frac{1}{6}(2y + 15)^{3/2} + \frac{15}{2}(2y + 15)^{1/2} = x + C

    And we know y(1) = -3.

    Thus
    \frac{1}{6}(2 \cdot -3 + 15)^{3/2} + \frac{15}{2}(2 \cdot -3 + 15)^{1/2} = 1 + C

    C = \frac{1}{6}(9)^{3/2} + \frac{15}{2}(9)^{1/2} - 1 = \frac{27}{6} + \frac{45}{2} - 1 = 26

    So finally:
    \frac{1}{6}(2y + 15)^{3/2} + \frac{15}{2}(2y + 15)^{1/2} = x + 26

    Now to solve for y.
    \frac{1}{6}(2y + 15)^{3/2} = -\frac{15}{2}(2y + 15)^{1/2} + x + 26 <-- Square both sides:

    \frac{1}{36}(2y + 15)^3 = \left ( -\frac{15}{2}(2y + 15)^{1/2} + x + 26 \right )^2

    \frac{1}{36}(2y + 15)^3 = x^2 - (15x + 390)\sqrt{2y + 15} + 52x + \frac{225y}{2} + \frac{6079}{4}

    Now solve this for \sqrt{2y + 15}, square both sides again, solve the polynomial for y, and check the solutions to make sure there are no extra ones in there. It'll keep you busy for a while, to say the least.

    -Dan
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