Question

Solve the initial value problem.

Answer 1

Separate the variables:

$y' = \dfrac{dy}{dx} = \dfrac{2x}{1+2y} \implies (1+2y) \, dy = 2x \, dx$

Integrate both sides:

$\displaystyle \int(1+2y) \, dy = \int 2x \, dx[/tex\[tex]y + y^2 = x^2 + C$

Use the given initial condition to solve for C :

$1 + 1^2 = 1^2 + C \implies C = 1$

So the particular solution is

$\boxed{y + y^2 = x^2 + 1}$

which you can also solve explicitly for y as a function of x. By completing the square on the left side, we have

$y + y^2 = x^2 + 1$

$\dfrac14 + y + y^2 = x^2 + \dfrac54$

$\left(\dfrac12 + y\right)^2 = x^2 + \dfrac54$

$\dfrac12 + y = \pm \sqrt{x^2+\dfrac54}$

Note that y(1) = 1 is positive, so the right side should involve the positive square root:

$\dfrac12 + y = \sqrt{x^2+\dfrac54}$

$\boxed{y = -\dfrac12 + \sqrt{x^2+\dfrac54} = -\dfrac12 + \dfrac12 \sqrt{4x^2+5}}$