Solve the separable differential equation y'(x) = sqrt(2y(x)+15)
and find the particular solution satisfying the initial condition y(1) = -3
Is $\displaystyle y^{\prime} = \sqrt{2y + 15}$?
Let me give you a hand:
$\displaystyle \int \frac{y}{\sqrt{2y + 15}}~dy$
Let $\displaystyle u = 2y + 15 \implies du = 2 dy$
$\displaystyle \int \frac{y}{\sqrt{2y + 15}}~dy = \int \frac{u - 15}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{dy}{2}$
$\displaystyle = \frac{1}{4} \int \left ( \sqrt{u} -
\frac{15}{\sqrt{u}} \right ) du$
Can you take it from here?
-Dan
Let me continue.
$\displaystyle = \frac{1}{4} \int \left ( \sqrt{u} -
\frac{15}{\sqrt{u}} \right ) du$
$\displaystyle = \frac{1}{4} \left ( \frac{2}{3}u^{3/2} + 2 \cdot 15u^{1/2} \right ) + C$
$\displaystyle = \frac{1}{6}(2y + 15)^{3/2} + \frac{15}{2}(2y + 15)^{1/2} + C$
So we have the differential equation:
$\displaystyle y^{\prime} = \sqrt{2y + 15}$
$\displaystyle \frac{dy}{\sqrt{2y + 15}} = dx$
$\displaystyle \int \frac{dy}{\sqrt{2y + 15}} = \int dx$
The LHS is just the integral I finished up at the top, so
$\displaystyle \frac{1}{6}(2y + 15)^{3/2} + \frac{15}{2}(2y + 15)^{1/2} = x + C$
And we know $\displaystyle y(1) = -3$.
Thus
$\displaystyle \frac{1}{6}(2 \cdot -3 + 15)^{3/2} + \frac{15}{2}(2 \cdot -3 + 15)^{1/2} = 1 + C$
$\displaystyle C = \frac{1}{6}(9)^{3/2} + \frac{15}{2}(9)^{1/2} - 1 = \frac{27}{6} + \frac{45}{2} - 1 = 26$
So finally:
$\displaystyle \frac{1}{6}(2y + 15)^{3/2} + \frac{15}{2}(2y + 15)^{1/2} = x + 26$
Now to solve for y.
$\displaystyle \frac{1}{6}(2y + 15)^{3/2} = -\frac{15}{2}(2y + 15)^{1/2} + x + 26$ <-- Square both sides:
$\displaystyle \frac{1}{36}(2y + 15)^3 = \left ( -\frac{15}{2}(2y + 15)^{1/2} + x + 26 \right )^2$
$\displaystyle \frac{1}{36}(2y + 15)^3 = x^2 - (15x + 390)\sqrt{2y + 15} + 52x + \frac{225y}{2} + \frac{6079}{4}$
Now solve this for $\displaystyle \sqrt{2y + 15}$, 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