Bernoulli Differential Equations — Question 9

Question 9 – Solution

Strategy. Periodicity removes the exponentially growing homogeneous term in the reciprocal variable; positivity must then be checked before inversion.

Step 1: Solve the linear equation. With $v=1/y$, the equation becomes $v'-v=-(2+\sin x)$. A particular solution and the general solution are $$v_p=2+\frac{\sin x+\cos x}{2},\qquad v=v_p+Ce^x.$$ Indeed $v_p'-v_p=-2-\sin x$.

Step 2: Select and bound the periodic solution. If positive $y$ is $2\pi$-periodic, then so is $v=1/y$. The term $Ce^x$ is periodic only if $C=0$. Since $$2-\frac{\sqrt 2}{2}\le v_p\le 2+\frac{\sqrt 2}{2},$$ its reciprocal exists globally and is positive. Therefore $$\boxed{y_p=\frac 1{2+(\sin x+\cos x)/2},\qquad y_p(0)=\frac 25}.$$ Its minimum is $1/(2+\sqrt 2/2)$ at $x=\pi/4$, and its maximum is $1/(2-\sqrt 2/2)$ at $x=5\pi/4$, modulo $2\pi$.

Step 3: Check the perturbed initial condition. For $y(0)=1/2$, $v(0)=2$, so $C=-1/2$. The function $v=v_p-e^x/2$ starts positive, but $v(2\pi)=5/2-e^{2\pi}/2<0$. It therefore has a first zero $T\in(0,2\pi)$. At this zero, $v'(T)=-(2+\sin T)<0$. Hence $v\downarrow 0$ from the positive side and $y=1/v\to+\infty$ as $x\uparrow T$.

Step 4: Verify the periodic branch. Since $v_p'=v_p-(2+\sin x)$, differentiation gives $y_p'=-y_p+(2+\sin x)y_p^2$, as required.