Separable Equations — Question 3

Question 3 – Solution

Strategy. Integration gives a squared relation; the initial condition and the excluded line determine which part is a solution.

Step 1: Integrate and choose the sign. $$(y-2)\,dy=-e^x\,dx,\qquad \frac 12(y-2)^2=-e^x+C.$$ At $(0,1)$, $C=3/2$, so $(y-2)^2=3-2e^x$. Since $y(0)-2=-1$, continuity selects $$\boxed{y=2-\sqrt{3-2e^x},\qquad I=(-\infty,\ln(3/2))}.$$ The radicand must be strictly positive: zero would give the forbidden value $y=2$.

Step 2: Check behavior and verify. Throughout $I$, $$y'=\frac{e^x}{\sqrt{3-2e^x}}=-\frac{e^x}{y-2}>0.$$ As $x\to-\infty$, $y\to 2-\sqrt 3$. As $x\uparrow\ln(3/2)$, $y\to 2$ but $y'\to+\infty$. A finite limit of $y$ alone does not permit extension: the equation is undefined at the endpoint and a finite derivative is impossible there.

Step 3: Audit the other branch. The branch $2+\sqrt{3-2e^x}$ satisfies the differential equation on $I$, but has $y(0)=3$, not $1$. Switching signs would still pass through $y=2$, and for larger $x$ the radicand is negative. Neither proposal extends the given real solution.