Intervals of Validity — Question 7

Question 7 – Solution

Strategy. A denominator zero is relevant only in the connected component containing the initial point. Preserve constant solutions before separating.

Step 1: Obtain a parameter formula. The constant solutions are $y=0,1$. Otherwise separation gives $\ln|y/(1-y)|=x+C$. Applying the initial value yields a formula that also includes both constants: $$\boxed{y(x)=\frac{pe^x}{1-p+pe^x}}.$$ Its denominator $D(x)$ satisfies $D(0)=1$. A zero requires $e^x=(p-1)/p>0$, which occurs exactly when $p<0$ or $p>1$.

Step 2: Select the component containing zero. When it exists, set $T=\ln((p-1)/p)$. The complete classification is $$\boxed{ \begin{array}{c|c|c} \text{initial value}&\text{endpoint location}&\text{maximal interval}\\ \hline p<0&T>0&(-\infty,T)\\ 0\le p\le 1&\text{no finite endpoint}&\mathbb R\\ p>1&T<0&(T,\infty) \end{array}}$$ For $0<p<1$, both terms of $D=1-p+pe^x$ are positive. The cases $p=0,1$ are global constants.

Step 3: Determine the endpoint behavior. If $p<0$, then $D\downarrow 0^+$ as $x\uparrow T$, while $pe^x\to p-1<0$, so $y\to-\infty$. If $p>1$, then $D\downarrow 0^+$ as $x\downarrow T$, while $pe^x\to p-1>0$, so $y\to+\infty$. Neither pole is removable.

Step 4: Verify and interpret. With $N=pe^x$, $N'=D'=N$, and $$y'=\frac{N(D-N)}{D^2}=y(1-y),\qquad y(0)=p.$$ Smoothness gives local existence and uniqueness. It does not prevent the finite-time blow-up found for $p\notin[0,1]$; the initial value determines whether the selected solution encounters a pole.