Bernoulli Differential Equations — Question 4

Question 4 – Solution

Strategy. The squared variable must be strictly positive, and the original reciprocal term remains undefined at zero.

Step 1: Linearize and integrate. Multiplication by $2y$ yields $v'+2v=2e^{-x}$, where $v=y^2$. Using the integrating factor $e^{2x}$, $$(e^{2x}v)'=2e^x,\qquad v=2e^{-x}+Ce^{-2x}.$$ Since $v(0)=1$, $C=-1$.

Step 2: Invert on the correct interval. The condition $v=e^{-2x}(2e^x-1)>0$ requires $x>-\ln 2$. The initial sign gives $$\boxed{y=-e^{-x}\sqrt{2e^x-1},\qquad I=(-\ln 2,\infty)}.$$ As $x\downarrow-\ln 2$, $y\to 0$ from below. The equation gives $y'=e^{-x}/y-y\to-\infty$, so no differentiable extension exists at that endpoint. Also the original right-hand side is undefined there. As $x\to\infty$, $y\to 0$ from below, but it never reaches zero at a finite point of $I$.

Step 3: Verify and check the excluded value. Differentiating $y^2=v$ gives $2yy'=v'$. Since $v'+2v=2e^{-x}$ and $y\ne 0$ on $I$, division by $2y$ recovers $y'+y=e^{-x}/y$. The selected formula gives $y(0)=-1$. Unlike positive-power examples, $\boxed{y\equiv 0\text{ is not a solution}}$: the original equation has no value at zero.