Bernoulli Differential Equations — Question 10

Question 10 – Solution

Strategy. Handle the exceptional equations directly, then analyze the apparent singularity of the general expression through its logarithm.

Step 1: Solve the elementary exceptional cases. If $n=0$, the equation is $y'+y=1$, giving $\boxed{y=1+e^{-x}}$. If $n=1$, the two $y$ terms cancel, giving $\boxed{y=2}$. Both are positive on $\mathbb R$; the first is linear and the second has zero derivative.

Step 2: Derive the parameterized formula. For $n\ne 1$, set $h=1-n\ne 0$ and $v=y^h$. Then $$v'+hv=h,\qquad v(0)=2^h,$$ so $v=1+(2^h-1)e^{-hx}$. Therefore $$\boxed{y_n(x)=\bigl[1+(2^{1-n}-1)e^{-(1-n)x}\bigr]^{1/(1-n)}}.$$ Use the connected interval through $0$ on which the bracket is strictly positive. This is necessary because the original positive solution has $v=y^h>0$; a negative base cannot be accepted by choosing a special rational exponent.

Step 3: Check the consistency conditions. At $n=0$, the formula becomes $1+e^{-x}$. At $x=0$, its bracket is $2^{1-n}$, and the positive-real power gives $2$. Conversely, differentiating $v=y^h$ on the stated interval recovers the original equation, so the inversion is valid there.

Step 4: Take the limit without dividing by zero. For fixed $x$, let $B(h)=1+(e^{h\ln 2}-1)e^{-hx}$. Since $B(0)=1$, it is positive for all sufficiently small $|h|$. Now $$\ln y_n=\frac{\ln B(h)}h,\qquad B'(0)=\ln 2.$$ The derivative definition gives $\lim_{h\to 0}\ln B(h)/h=\ln 2$. Exponentiating yields $\boxed{\lim_{n\to 1}y_n(x)=2}$, exactly the direct $n=1$ solution. Joint continuity on the compact segment between $0$ and the fixed $x$ keeps the bracket positive there for small $|h|$, so these values belong to the IVP branch.