Bernoulli Differential Equations — Question 2

Question 2 – Solution

Strategy. The derivative of $y^{-2}$ carries a negative factor, and the inverse transformation needs a separately chosen sign.

Step 1: Correct the transformed equation. For $y\ne 0$, divide by $y^3$ and use $v'=-2y^{-3}y'$. This gives $$-\frac 12v'+v=e^{2x},\qquad \boxed{v'-2v=-2e^{2x}}.$$ The student’s two coefficient signs are reversed. Moreover, $v=1/y^2$ is the same for $y$ and $-y$, so selecting the positive inverse would contradict the negative initial value.

Step 2: Solve and choose the branch. The integrating factor is $e^{-2x}$: $$(e^{-2x}v)'=-2,\qquad v=e^{2x}(C-2x).$$ The condition $v(0)=1$ gives $C=1$. Continuity and $y(0)=-1$ select $$\boxed{y=-\frac{e^{-x}}{\sqrt{1-2x}},\qquad I=(-\infty,1/2)}.$$ Here $v>0$. At $x\uparrow 1/2$, the solution tends to $-\infty$, so there is no finite extension; for $x>1/2$, this $v$ cannot equal $1/y^2$ for real $y$.

Step 3: Verify and distinguish other solutions. On $I$, logarithmic differentiation of $|y|$ gives $$\frac{y'}y=-1+\frac 1{1-2x}.$$ Consequently $y'+y=y/(1-2x)=e^{2x}y^3$, and the initial value is $-1$. The opposite branch $+e^{-x}/\sqrt{1-2x}$ also solves the differential equation, with initial value $1$. The zero function solves it as well but is omitted by $y^{-2}$. Neither of those alternatives solves the stated IVP.