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Find an explicit solution of an initial value problem $$\frac{dy}{dx}=\frac{2x}{1+2y}, \; y(2)=0$$

Attempt: I have no problem finding the general solution which is

$y+y^2=x^2+C$

Then, I find the implict solution which I think is correct, but I am not sure

$y+y^2=x^2-4$

Now, how should I go about finding an explicit solution? Thanks for your help.

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You want it to be $y + y^2 = x^2 - 4$, since you want $(x,y) = (2,0)$ to be a solution. –  Christopher A. Wong Feb 6 '13 at 21:26
    
@ChristopherA.Wong yep. Corrected –  Koba Feb 6 '13 at 21:33
    
Arhhh answers so good. Which one to accept? –  Koba Feb 6 '13 at 21:35

3 Answers 3

up vote 2 down vote accepted

Solve using the quadratic formula:

$$y^2+y-(x^2-4)=0 \implies y = \frac{-1 \pm \sqrt{1 + 4(x^2-4)}}{2}$$

You should choose the $+$ branch to satisfy the initial condition. Thus:

$$y(x) = \frac{\sqrt{4 x^2-15}-1}{2}$$

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We have $C=-4$. Now solve the quadratic equation $y^2+y-x^2+4=0$ for $y$. Use the Quadratic Formula.

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Complete the square on the right hand side i.e. \begin{align} y+y^2 & = x^2-4\\ y^2 + y + \dfrac14 & = x^2-4 + \dfrac14\\ \left( y + \dfrac12 \right)^2 & = x^2-\dfrac{15}4\\ y + \dfrac12 & = \pm \sqrt{x^2 - \dfrac{15}4}\\ y & = - \dfrac12 \pm \dfrac{\sqrt{4x^2-15}}2 \end{align} Further, we need $y(2) = 0$ and hence $y = - \dfrac12 + \dfrac{\sqrt{4x^2-15}}2$

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