$y'=(8\cos8x)/(3+2y)$ with $y(0)=-1$ Initial Value Problem: DiffEq

Question

Im told to find the explicit form $y(x)$ from the given differential equation and its initial value. Then find where the solution $x=?$ attain a maximum. What I did was:

$3+2ydy=8\cos 8xdx$ integrated both sides

$3y+y^2=\sin8x+c $

Im stuck because I can't figure out how to get $y$ by itself.

Answer 1

HINT: Use the quadratic formula, (Pretend the expression in $x$ is constant). Also, at the point you are at now, you can solve for the constant $c$, using your initial conditions.

Answer 2

To find where the function achieves its maximum, just put $y'=0$ in the differential equation

$$ y'=\frac{8\cos(8x)}{y+2} \implies 0=\frac{8\cos(8x)}{y+2} \implies \cos(8x)=0\implies \dots. $$

Answer 3

I often prefer completing the square to using the explicit quadratic formula, even though they are equivalent.

Since $(y+b)^2 = y^2 + 2by + b^2$, to complete the square for $y^2+3y$, $2b=3$ or $b = 3/2$. To complete the square add $(3/2)^2 = 9/4$ to both sides getting

$$ \sin 8x+c +9/4 = y^2+3y+9/4 = (y+3/2)^2.$$

You can now easily solve for $y$ and use the initial conditions.