Separable Equations — Question 8

Question 8 – Solution

Strategy. Track the denominator of the separated formula throughout an interval, not merely at its endpoints.

Step 1: Integrate. Since $a>0$, separation is valid near $0$: $$-\frac 1y=-\cos x+C,\qquad \boxed{y=\frac{a}{1-a+a\cos x}}.$$ Its derivative is $a^2\sin x/(1-a+a\cos x)^2=(\sin x)y^2$, and $y(0)=a$.

Step 2: Classify all positive parameters. The denominator $D=1-a+a\cos x$ ranges from $1-2a$ to $1$.

• If $0<a<1/2$, then $D>0$ everywhere: $I=\mathbb R$ and $y$ is bounded.

• If $a=1/2$, then $y=1/(1+\cos x)$ and $I=(-\pi,\pi)$.

• If $a>1/2$, let $\alpha=\arccos(1-1/a)\in(0,\pi)$. Then $I=(-\alpha,\alpha)$.

In the last two cases $D\downarrow 0$ at either endpoint from inside $I$, so $y\to+\infty$. Thus the threshold itself already blows up.

Step 3: Determine extrema and audit the return claim. For $0<a<1/2$, $\min y=a$ at $x=2n\pi$ and $\max y=a/(1-2a)$ at $x=(2n+1)\pi$, $n\in\mathbb Z$; $2\pi$ is a period. Only this parameter range gives a solution through the full cycle. For $a\ge 1/2$, a pole occurs before reaching $2\pi$. Substituting $2\pi$ into the algebraic formula gives a value on a disconnected branch, not a continuation of the initial solution.